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

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

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(X)

Q is empty.

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

MARK1(half1(X)) -> A__HALF1(mark1(X))
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
A__SQR1(s1(X)) -> MARK1(X)
MARK1(sqr1(X)) -> A__SQR1(mark1(X))
MARK1(dbl1(X)) -> A__DBL1(mark1(X))
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(recip1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
A__SQR1(s1(X)) -> A__SQR1(mark1(X))
MARK1(terms1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
A__ADD2(s1(X), Y) -> A__ADD2(mark1(X), mark1(Y))
A__DBL1(s1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(dbl1(X)) -> MARK1(X)
A__HALF1(s1(s1(X))) -> A__HALF1(mark1(X))
MARK1(sqr1(X)) -> MARK1(X)
A__SQR1(s1(X)) -> A__ADD2(a__sqr1(mark1(X)), a__dbl1(mark1(X)))
MARK1(half1(X)) -> MARK1(X)
A__ADD2(s1(X), Y) -> MARK1(X)
A__HALF1(dbl1(X)) -> MARK1(X)
A__TERMS1(N) -> MARK1(N)
MARK1(s1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
A__HALF1(s1(s1(X))) -> MARK1(X)
A__ADD2(s1(X), Y) -> MARK1(Y)
A__DBL1(s1(X)) -> A__DBL1(mark1(X))
A__TERMS1(N) -> A__SQR1(mark1(N))
A__SQR1(s1(X)) -> A__DBL1(mark1(X))

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(X)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

MARK1(half1(X)) -> A__HALF1(mark1(X))
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
A__SQR1(s1(X)) -> MARK1(X)
MARK1(sqr1(X)) -> A__SQR1(mark1(X))
MARK1(dbl1(X)) -> A__DBL1(mark1(X))
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(recip1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
A__SQR1(s1(X)) -> A__SQR1(mark1(X))
MARK1(terms1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
A__ADD2(s1(X), Y) -> A__ADD2(mark1(X), mark1(Y))
A__DBL1(s1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(dbl1(X)) -> MARK1(X)
A__HALF1(s1(s1(X))) -> A__HALF1(mark1(X))
MARK1(sqr1(X)) -> MARK1(X)
A__SQR1(s1(X)) -> A__ADD2(a__sqr1(mark1(X)), a__dbl1(mark1(X)))
MARK1(half1(X)) -> MARK1(X)
A__ADD2(s1(X), Y) -> MARK1(X)
A__HALF1(dbl1(X)) -> MARK1(X)
A__TERMS1(N) -> MARK1(N)
MARK1(s1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
A__HALF1(s1(s1(X))) -> MARK1(X)
A__ADD2(s1(X), Y) -> MARK1(Y)
A__DBL1(s1(X)) -> A__DBL1(mark1(X))
A__TERMS1(N) -> A__SQR1(mark1(N))
A__SQR1(s1(X)) -> A__DBL1(mark1(X))

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(X)

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


The following pairs can be strictly oriented and are deleted.


MARK1(half1(X)) -> A__HALF1(mark1(X))
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
A__SQR1(s1(X)) -> MARK1(X)
MARK1(sqr1(X)) -> A__SQR1(mark1(X))
MARK1(dbl1(X)) -> A__DBL1(mark1(X))
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(first2(X1, X2)) -> MARK1(X2)
A__SQR1(s1(X)) -> A__SQR1(mark1(X))
MARK1(terms1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
MARK1(terms1(X)) -> A__TERMS1(mark1(X))
A__ADD2(s1(X), Y) -> A__ADD2(mark1(X), mark1(Y))
A__DBL1(s1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(dbl1(X)) -> MARK1(X)
A__HALF1(s1(s1(X))) -> A__HALF1(mark1(X))
MARK1(sqr1(X)) -> MARK1(X)
A__SQR1(s1(X)) -> A__ADD2(a__sqr1(mark1(X)), a__dbl1(mark1(X)))
MARK1(half1(X)) -> MARK1(X)
A__ADD2(s1(X), Y) -> MARK1(X)
A__HALF1(dbl1(X)) -> MARK1(X)
A__TERMS1(N) -> MARK1(N)
MARK1(s1(X)) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
A__HALF1(s1(s1(X))) -> MARK1(X)
A__ADD2(s1(X), Y) -> MARK1(Y)
A__DBL1(s1(X)) -> A__DBL1(mark1(X))
A__SQR1(s1(X)) -> A__DBL1(mark1(X))
The remaining pairs can at least by weakly be oriented.

MARK1(recip1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__TERMS1(N) -> A__SQR1(mark1(N))
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
half1(x1)  =  half1(x1)
A__HALF1(x1)  =  A__HALF1(x1)
mark1(x1)  =  x1
first2(x1, x2)  =  first2(x1, x2)
A__FIRST2(x1, x2)  =  A__FIRST1(x2)
A__SQR1(x1)  =  A__SQR1(x1)
s1(x1)  =  s1(x1)
sqr1(x1)  =  sqr1(x1)
dbl1(x1)  =  dbl1(x1)
A__DBL1(x1)  =  A__DBL1(x1)
add2(x1, x2)  =  add2(x1, x2)
recip1(x1)  =  x1
cons2(x1, x2)  =  x1
terms1(x1)  =  terms1(x1)
A__ADD2(x1, x2)  =  A__ADD2(x1, x2)
0  =  0
A__TERMS1(x1)  =  A__TERMS1(x1)
a__sqr1(x1)  =  a__sqr1(x1)
a__dbl1(x1)  =  a__dbl1(x1)
a__half1(x1)  =  a__half1(x1)
a__first2(x1, x2)  =  a__first2(x1, x2)
a__add2(x1, x2)  =  a__add2(x1, x2)
a__terms1(x1)  =  a__terms1(x1)
nil  =  nil

Lexicographic Path Order [19].
Precedence:
[first2, afirst2, nil] > [MARK1, half1, AHALF1, AFIRST1, ahalf1] > s1
[first2, afirst2, nil] > [MARK1, half1, AHALF1, AFIRST1, ahalf1] > 0
[ASQR1, sqr1, dbl1, ADBL1, terms1, ATERMS1, asqr1, adbl1, aterms1] > [add2, AADD2, aadd2] > [MARK1, half1, AHALF1, AFIRST1, ahalf1] > s1
[ASQR1, sqr1, dbl1, ADBL1, terms1, ATERMS1, asqr1, adbl1, aterms1] > [add2, AADD2, aadd2] > [MARK1, half1, AHALF1, AFIRST1, ahalf1] > 0


The following usable rules [14] were oriented:

mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
a__add2(0, X) -> mark1(X)
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
a__half1(dbl1(X)) -> mark1(X)
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__first2(X1, X2) -> first2(X1, X2)
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__dbl1(X) -> dbl1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__add2(X1, X2) -> add2(X1, X2)
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(X) -> half1(X)
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__sqr1(X) -> sqr1(X)
a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__terms1(X) -> terms1(X)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

A__TERMS1(N) -> A__SQR1(mark1(N))
MARK1(recip1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

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

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

MARK1(recip1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(X)

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


The following pairs can be strictly oriented and are deleted.


MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least by weakly be oriented.

MARK1(recip1(X)) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
recip1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

MARK1(recip1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(X)

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


The following pairs can be strictly oriented and are deleted.


MARK1(recip1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
recip1(x1)  =  recip1(x1)

Lexicographic Path Order [19].
Precedence:
recip1 > MARK1


The following usable rules [14] were oriented: none



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

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

a__terms1(N) -> cons2(recip1(a__sqr1(mark1(N))), terms1(s1(N)))
a__sqr1(0) -> 0
a__sqr1(s1(X)) -> s1(a__add2(a__sqr1(mark1(X)), a__dbl1(mark1(X))))
a__dbl1(0) -> 0
a__dbl1(s1(X)) -> s1(s1(a__dbl1(mark1(X))))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(a__add2(mark1(X), mark1(Y)))
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__half1(0) -> 0
a__half1(s1(0)) -> 0
a__half1(s1(s1(X))) -> s1(a__half1(mark1(X)))
a__half1(dbl1(X)) -> mark1(X)
mark1(terms1(X)) -> a__terms1(mark1(X))
mark1(sqr1(X)) -> a__sqr1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(dbl1(X)) -> a__dbl1(mark1(X))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(half1(X)) -> a__half1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(recip1(X)) -> recip1(mark1(X))
mark1(s1(X)) -> s1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
a__terms1(X) -> terms1(X)
a__sqr1(X) -> sqr1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__dbl1(X) -> dbl1(X)
a__first2(X1, X2) -> first2(X1, X2)
a__half1(X) -> half1(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.