Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
RECIP1(mark1(X)) -> RECIP1(X)
ACTIVE1(recip1(X)) -> RECIP1(active1(X))
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
RECIP1(ok1(X)) -> RECIP1(X)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(dbl1(s1(X))) -> DBL1(X)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(terms1(X)) -> ACTIVE1(X)
ACTIVE1(terms1(N)) -> SQR1(N)
ACTIVE1(first2(X1, X2)) -> FIRST2(active1(X1), X2)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(sqr1(X)) -> PROPER1(X)
ACTIVE1(dbl1(X)) -> ACTIVE1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(add2(X1, X2)) -> ADD2(active1(X1), X2)
SQR1(mark1(X)) -> SQR1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(terms1(N)) -> CONS2(recip1(sqr1(N)), terms1(s1(N)))
PROPER1(sqr1(X)) -> SQR1(proper1(X))
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(add2(X1, X2)) -> ADD2(X1, active1(X2))
PROPER1(recip1(X)) -> RECIP1(proper1(X))
ACTIVE1(add2(s1(X), Y)) -> ADD2(X, Y)
ACTIVE1(sqr1(s1(X))) -> ADD2(sqr1(X), dbl1(X))
ACTIVE1(terms1(N)) -> RECIP1(sqr1(N))
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> FIRST2(X, Z)
TOP1(mark1(X)) -> PROPER1(X)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
PROPER1(terms1(X)) -> PROPER1(X)
DBL1(ok1(X)) -> DBL1(X)
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)
ACTIVE1(sqr1(s1(X))) -> DBL1(X)
ACTIVE1(first2(X1, X2)) -> FIRST2(X1, active1(X2))
ACTIVE1(dbl1(X)) -> DBL1(active1(X))
PROPER1(add2(X1, X2)) -> PROPER1(X1)
PROPER1(add2(X1, X2)) -> ADD2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> CONS2(Y, first2(X, Z))
PROPER1(recip1(X)) -> PROPER1(X)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(sqr1(s1(X))) -> S1(add2(sqr1(X), dbl1(X)))
TERMS1(ok1(X)) -> TERMS1(X)
ACTIVE1(terms1(X)) -> TERMS1(active1(X))
ACTIVE1(sqr1(X)) -> SQR1(active1(X))
ACTIVE1(terms1(N)) -> TERMS1(s1(N))
DBL1(mark1(X)) -> DBL1(X)
PROPER1(first2(X1, X2)) -> FIRST2(proper1(X1), proper1(X2))
ACTIVE1(dbl1(s1(X))) -> S1(dbl1(X))
PROPER1(add2(X1, X2)) -> PROPER1(X2)
ACTIVE1(sqr1(s1(X))) -> SQR1(X)
FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
PROPER1(dbl1(X)) -> DBL1(proper1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
SQR1(ok1(X)) -> SQR1(X)
TERMS1(mark1(X)) -> TERMS1(X)
PROPER1(terms1(X)) -> TERMS1(proper1(X))
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(terms1(N)) -> S1(N)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(dbl1(s1(X))) -> S1(s1(dbl1(X)))
PROPER1(dbl1(X)) -> PROPER1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)
ACTIVE1(add2(s1(X), Y)) -> S1(add2(X, Y))

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
RECIP1(mark1(X)) -> RECIP1(X)
ACTIVE1(recip1(X)) -> RECIP1(active1(X))
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
RECIP1(ok1(X)) -> RECIP1(X)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(dbl1(s1(X))) -> DBL1(X)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(terms1(X)) -> ACTIVE1(X)
ACTIVE1(terms1(N)) -> SQR1(N)
ACTIVE1(first2(X1, X2)) -> FIRST2(active1(X1), X2)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(sqr1(X)) -> PROPER1(X)
ACTIVE1(dbl1(X)) -> ACTIVE1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(add2(X1, X2)) -> ADD2(active1(X1), X2)
SQR1(mark1(X)) -> SQR1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(terms1(N)) -> CONS2(recip1(sqr1(N)), terms1(s1(N)))
PROPER1(sqr1(X)) -> SQR1(proper1(X))
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(add2(X1, X2)) -> ADD2(X1, active1(X2))
PROPER1(recip1(X)) -> RECIP1(proper1(X))
ACTIVE1(add2(s1(X), Y)) -> ADD2(X, Y)
ACTIVE1(sqr1(s1(X))) -> ADD2(sqr1(X), dbl1(X))
ACTIVE1(terms1(N)) -> RECIP1(sqr1(N))
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> FIRST2(X, Z)
TOP1(mark1(X)) -> PROPER1(X)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
PROPER1(terms1(X)) -> PROPER1(X)
DBL1(ok1(X)) -> DBL1(X)
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)
ACTIVE1(sqr1(s1(X))) -> DBL1(X)
ACTIVE1(first2(X1, X2)) -> FIRST2(X1, active1(X2))
ACTIVE1(dbl1(X)) -> DBL1(active1(X))
PROPER1(add2(X1, X2)) -> PROPER1(X1)
PROPER1(add2(X1, X2)) -> ADD2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> CONS2(Y, first2(X, Z))
PROPER1(recip1(X)) -> PROPER1(X)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(sqr1(s1(X))) -> S1(add2(sqr1(X), dbl1(X)))
TERMS1(ok1(X)) -> TERMS1(X)
ACTIVE1(terms1(X)) -> TERMS1(active1(X))
ACTIVE1(sqr1(X)) -> SQR1(active1(X))
ACTIVE1(terms1(N)) -> TERMS1(s1(N))
DBL1(mark1(X)) -> DBL1(X)
PROPER1(first2(X1, X2)) -> FIRST2(proper1(X1), proper1(X2))
ACTIVE1(dbl1(s1(X))) -> S1(dbl1(X))
PROPER1(add2(X1, X2)) -> PROPER1(X2)
ACTIVE1(sqr1(s1(X))) -> SQR1(X)
FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
PROPER1(dbl1(X)) -> DBL1(proper1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
SQR1(ok1(X)) -> SQR1(X)
TERMS1(mark1(X)) -> TERMS1(X)
PROPER1(terms1(X)) -> TERMS1(proper1(X))
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(terms1(N)) -> S1(N)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(dbl1(s1(X))) -> S1(s1(dbl1(X)))
PROPER1(dbl1(X)) -> PROPER1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)
ACTIVE1(add2(s1(X), Y)) -> S1(add2(X, Y))

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 11 SCCs with 35 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

S1(ok1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


S1(ok1(X)) -> S1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
S1(x1)  =  S1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > S1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
The remaining pairs can at least by weakly be oriented.

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
Used ordering: Combined order from the following AFS and order.
FIRST2(x1, x2)  =  x2
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
The remaining pairs can at least by weakly be oriented.

FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
Used ordering: Combined order from the following AFS and order.
FIRST2(x1, x2)  =  FIRST1(x2)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[FIRST1, mark1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FIRST2(x1, x2)  =  FIRST1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[FIRST1, mark1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DBL1(mark1(X)) -> DBL1(X)
DBL1(ok1(X)) -> DBL1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DBL1(ok1(X)) -> DBL1(X)
The remaining pairs can at least by weakly be oriented.

DBL1(mark1(X)) -> DBL1(X)
Used ordering: Combined order from the following AFS and order.
DBL1(x1)  =  DBL1(x1)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

DBL1(mark1(X)) -> DBL1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


DBL1(mark1(X)) -> DBL1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DBL1(x1)  =  DBL1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
mark1 > DBL1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)
The remaining pairs can at least by weakly be oriented.

ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
Used ordering: Combined order from the following AFS and order.
ADD2(x1, x2)  =  x2
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
The remaining pairs can at least by weakly be oriented.

ADD2(mark1(X1), X2) -> ADD2(X1, X2)
Used ordering: Combined order from the following AFS and order.
ADD2(x1, x2)  =  ADD1(x2)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[ADD1, mark1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ADD2(mark1(X1), X2) -> ADD2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ADD2(mark1(X1), X2) -> ADD2(X1, X2)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADD2(x1, x2)  =  ADD1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[ADD1, mark1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SQR1(ok1(X)) -> SQR1(X)
SQR1(mark1(X)) -> SQR1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


SQR1(mark1(X)) -> SQR1(X)
The remaining pairs can at least by weakly be oriented.

SQR1(ok1(X)) -> SQR1(X)
Used ordering: Combined order from the following AFS and order.
SQR1(x1)  =  SQR1(x1)
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SQR1(ok1(X)) -> SQR1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


SQR1(ok1(X)) -> SQR1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SQR1(x1)  =  SQR1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > SQR1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

RECIP1(mark1(X)) -> RECIP1(X)
RECIP1(ok1(X)) -> RECIP1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


RECIP1(ok1(X)) -> RECIP1(X)
The remaining pairs can at least by weakly be oriented.

RECIP1(mark1(X)) -> RECIP1(X)
Used ordering: Combined order from the following AFS and order.
RECIP1(x1)  =  RECIP1(x1)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

RECIP1(mark1(X)) -> RECIP1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


RECIP1(mark1(X)) -> RECIP1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
RECIP1(x1)  =  RECIP1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
mark1 > RECIP1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The remaining pairs can at least by weakly be oriented.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used ordering: Combined order from the following AFS and order.
CONS2(x1, x2)  =  x2
mark1(x1)  =  mark
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

CONS2(mark1(X1), X2) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


CONS2(mark1(X1), X2) -> CONS2(X1, X2)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CONS2(x1, x2)  =  CONS1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[CONS1, mark1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TERMS1(mark1(X)) -> TERMS1(X)
TERMS1(ok1(X)) -> TERMS1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TERMS1(ok1(X)) -> TERMS1(X)
The remaining pairs can at least by weakly be oriented.

TERMS1(mark1(X)) -> TERMS1(X)
Used ordering: Combined order from the following AFS and order.
TERMS1(x1)  =  TERMS1(x1)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TERMS1(mark1(X)) -> TERMS1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TERMS1(mark1(X)) -> TERMS1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TERMS1(x1)  =  TERMS1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
mark1 > TERMS1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(add2(X1, X2)) -> PROPER1(X1)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(terms1(X)) -> PROPER1(X)
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)
PROPER1(dbl1(X)) -> PROPER1(X)
PROPER1(add2(X1, X2)) -> PROPER1(X2)
PROPER1(sqr1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(add2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)
PROPER1(add2(X1, X2)) -> PROPER1(X2)
The remaining pairs can at least by weakly be oriented.

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(terms1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)
PROPER1(sqr1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  x1
recip1(x1)  =  x1
add2(x1, x2)  =  add2(x1, x2)
s1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
terms1(x1)  =  x1
first2(x1, x2)  =  first2(x1, x2)
dbl1(x1)  =  x1
sqr1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(terms1(X)) -> PROPER1(X)
PROPER1(sqr1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(s1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(terms1(X)) -> PROPER1(X)
PROPER1(sqr1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
recip1(x1)  =  x1
s1(x1)  =  s1(x1)
terms1(x1)  =  x1
sqr1(x1)  =  x1
dbl1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(terms1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)
PROPER1(sqr1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(terms1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)
PROPER1(sqr1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
recip1(x1)  =  x1
terms1(x1)  =  terms1(x1)
dbl1(x1)  =  x1
sqr1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(sqr1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(sqr1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
recip1(x1)  =  x1
sqr1(x1)  =  sqr1(x1)
dbl1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[PROPER1, sqr1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(recip1(X)) -> PROPER1(X)
PROPER1(dbl1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(dbl1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(recip1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
recip1(x1)  =  x1
dbl1(x1)  =  dbl1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(recip1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(recip1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
recip1(x1)  =  recip1(x1)

Lexicographic Path Order [19].
Precedence:
recip1 > PROPER1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
QDP
                                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(terms1(X)) -> ACTIVE1(X)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(terms1(X)) -> ACTIVE1(X)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  x1
dbl1(x1)  =  x1
first2(x1, x2)  =  first2(x1, x2)
cons2(x1, x2)  =  x1
terms1(x1)  =  x1
add2(x1, x2)  =  add2(x1, x2)
sqr1(x1)  =  x1
recip1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(terms1(X)) -> ACTIVE1(X)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(terms1(X)) -> ACTIVE1(X)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
dbl1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
terms1(x1)  =  x1
sqr1(x1)  =  x1
recip1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(terms1(X)) -> ACTIVE1(X)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(terms1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
dbl1(x1)  =  x1
terms1(x1)  =  terms1(x1)
sqr1(x1)  =  x1
recip1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(sqr1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(sqr1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
dbl1(x1)  =  x1
sqr1(x1)  =  sqr1(x1)
recip1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[ACTIVE1, sqr1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
ACTIVE1(recip1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(recip1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(dbl1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
dbl1(x1)  =  x1
recip1(x1)  =  recip1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(dbl1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(dbl1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
dbl1(x1)  =  dbl1(x1)

Lexicographic Path Order [19].
Precedence:
dbl1 > ACTIVE1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ QDPOrderProof
QDP
                                    ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TOP1(mark1(X)) -> TOP1(proper1(X))
The remaining pairs can at least by weakly be oriented.

TOP1(ok1(X)) -> TOP1(active1(X))
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  x1
ok1(x1)  =  x1
active1(x1)  =  x1
mark1(x1)  =  mark1(x1)
proper1(x1)  =  x1
terms1(x1)  =  terms1(x1)
cons2(x1, x2)  =  cons1(x1)
recip1(x1)  =  recip1(x1)
sqr1(x1)  =  sqr1(x1)
s1(x1)  =  s
0  =  0
add2(x1, x2)  =  add2(x1, x2)
dbl1(x1)  =  dbl1(x1)
first2(x1, x2)  =  first2(x1, x2)
nil  =  nil

Lexicographic Path Order [19].
Precedence:
terms1 > recip1 > [mark1, cons1]
terms1 > sqr1 > s > add2 > [mark1, cons1]
terms1 > sqr1 > 0 > [mark1, cons1]
terms1 > sqr1 > 0 > nil
dbl1 > s > add2 > [mark1, cons1]
dbl1 > 0 > [mark1, cons1]
dbl1 > 0 > nil
first2 > [mark1, cons1]
first2 > nil


The following usable rules [14] were oriented:

proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
dbl1(ok1(X)) -> ok1(dbl1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
recip1(mark1(X)) -> mark1(recip1(X))
recip1(ok1(X)) -> ok1(recip1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
terms1(mark1(X)) -> mark1(terms1(X))
terms1(ok1(X)) -> ok1(terms1(X))
active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

TOP1(ok1(X)) -> TOP1(active1(X))

The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TOP1(ok1(X)) -> TOP1(active1(X))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TOP1(x1)  =  TOP1(x1)
ok1(x1)  =  ok1(x1)
active1(x1)  =  x1
terms1(x1)  =  x1
mark1(x1)  =  x1
cons2(x1, x2)  =  x2
recip1(x1)  =  x1
sqr1(x1)  =  x1
s1(x1)  =  x1
0  =  0
add2(x1, x2)  =  x2
dbl1(x1)  =  x1
first2(x1, x2)  =  x1
nil  =  nil

Lexicographic Path Order [19].
Precedence:
0 > nil


The following usable rules [14] were oriented:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
dbl1(ok1(X)) -> ok1(dbl1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
sqr1(mark1(X)) -> mark1(sqr1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
recip1(mark1(X)) -> mark1(recip1(X))
recip1(ok1(X)) -> ok1(recip1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
terms1(mark1(X)) -> mark1(terms1(X))
terms1(ok1(X)) -> ok1(terms1(X))
s1(ok1(X)) -> ok1(s1(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(terms1(N)) -> mark1(cons2(recip1(sqr1(N)), terms1(s1(N))))
active1(sqr1(0)) -> mark1(0)
active1(sqr1(s1(X))) -> mark1(s1(add2(sqr1(X), dbl1(X))))
active1(dbl1(0)) -> mark1(0)
active1(dbl1(s1(X))) -> mark1(s1(s1(dbl1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(terms1(X)) -> terms1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(recip1(X)) -> recip1(active1(X))
active1(sqr1(X)) -> sqr1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(dbl1(X)) -> dbl1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
terms1(mark1(X)) -> mark1(terms1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
recip1(mark1(X)) -> mark1(recip1(X))
sqr1(mark1(X)) -> mark1(sqr1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
dbl1(mark1(X)) -> mark1(dbl1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
proper1(terms1(X)) -> terms1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(recip1(X)) -> recip1(proper1(X))
proper1(sqr1(X)) -> sqr1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(0) -> ok1(0)
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(dbl1(X)) -> dbl1(proper1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
terms1(ok1(X)) -> ok1(terms1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
recip1(ok1(X)) -> ok1(recip1(X))
sqr1(ok1(X)) -> ok1(sqr1(X))
s1(ok1(X)) -> ok1(s1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
dbl1(ok1(X)) -> ok1(dbl1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.