Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx26_Luc03b

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.

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 contains 11 SCCs with 35 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FIRST₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FIRST₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FIRST₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DBL₁(ok₁(X)) -> DBL₁(X)

Used argument filtering: DBL₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DBL₁(mark₁(X)) -> DBL₁(X)

Used argument filtering: DBL₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: ADD₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SQR₁(mark₁(X)) -> SQR₁(X)

Used argument filtering: SQR₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SQR₁(ok₁(X)) -> SQR₁(X)

Used argument filtering: SQR₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

RECIP₁(ok₁(X)) -> RECIP₁(X)

Used argument filtering: RECIP₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

RECIP₁(mark₁(X)) -> RECIP₁(X)

Used argument filtering: RECIP₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: CONS₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TERMS₁(ok₁(X)) -> TERMS₁(X)

Used argument filtering: TERMS₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TERMS₁(mark₁(X)) -> TERMS₁(X)

Used argument filtering: TERMS₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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)

Used argument filtering: 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
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(dbl₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
recip₁(x1) = x1
s₁(x1) = x1
terms₁(x1) = x1
sqr₁(x1) = x1
dbl₁(x1) = dbl₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(sqr₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
recip₁(x1) = x1
s₁(x1) = x1
terms₁(x1) = x1
sqr₁(x1) = sqr₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(terms₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
recip₁(x1) = x1
s₁(x1) = x1
terms₁(x1) = terms₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(recip₁(X)) -> PROPER₁(X)
PROPER₁(s₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
recip₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(recip₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
recip₁(x1) = recip₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

                                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(first₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)

Used argument filtering: 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
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(recip₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
dbl₁(x1) = x1
cons₂(x1, x2) = x1
terms₁(x1) = x1
sqr₁(x1) = x1
recip₁(x1) = recip₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(sqr₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
dbl₁(x1) = x1
cons₂(x1, x2) = x1
terms₁(x1) = x1
sqr₁(x1) = sqr₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(terms₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
dbl₁(x1) = x1
cons₂(x1, x2) = x1
terms₁(x1) = terms₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
dbl₁(x1) = x1
cons₂(x1, x2) = cons₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(dbl₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
dbl₁(x1) = dbl₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

                                  ↳ 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

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.