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__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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:

MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(dbl(X)) → A__DBL(mark(X))
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → MARK(X2)
MARK(first(X1, X2)) → MARK(X2)
MARK(recip(X)) → MARK(X)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(terms(X)) → MARK(X)
MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
A__TERMS(N) → MARK(N)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(add(X1, X2)) → MARK(X1)
A__TERMS(N) → A__SQR(mark(N))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(dbl(X)) → A__DBL(mark(X))
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → MARK(X2)
MARK(first(X1, X2)) → MARK(X2)
MARK(recip(X)) → MARK(X)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(terms(X)) → MARK(X)
MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
A__TERMS(N) → MARK(N)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(add(X1, X2)) → MARK(X1)
A__TERMS(N) → A__SQR(mark(N))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(sqr(X)) → A__SQR(mark(X))
MARK(dbl(X)) → A__DBL(mark(X))
MARK(add(X1, X2)) → MARK(X2)
MARK(sqr(X)) → MARK(X)
MARK(first(X1, X2)) → MARK(X2)
MARK(recip(X)) → MARK(X)
MARK(terms(X)) → MARK(X)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
A__TERMS(N) → MARK(N)
MARK(add(X1, X2)) → MARK(X1)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__TERMS(N) → A__SQR(mark(N))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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 3 less nodes.

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

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

MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → MARK(X2)
MARK(first(X1, X2)) → MARK(X2)
MARK(recip(X)) → MARK(X)
MARK(terms(X)) → MARK(X)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
A__TERMS(N) → MARK(N)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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 oriented strictly and are deleted.


MARK(add(X1, X2)) → MARK(X2)
MARK(first(X1, X2)) → MARK(X2)
MARK(terms(X)) → MARK(X)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
The remaining pairs can at least be oriented weakly.

MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(sqr(X)) → MARK(X)
MARK(recip(X)) → MARK(X)
A__TERMS(N) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
dbl(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
A__FIRST(x1, x2)  =  A__FIRST(x1, x2)
mark(x1)  =  x1
sqr(x1)  =  x1
add(x1, x2)  =  add(x1, x2)
recip(x1)  =  x1
terms(x1)  =  terms(x1)
s(x1)  =  s
cons(x1, x2)  =  x1
A__ADD(x1, x2)  =  A__ADD(x2)
0  =  0
A__TERMS(x1)  =  x1
a__sqr(x1)  =  x1
a__add(x1, x2)  =  a__add(x1, x2)
a__first(x1, x2)  =  a__first(x1, x2)
nil  =  nil
a__dbl(x1)  =  x1
a__terms(x1)  =  a__terms(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
[first2, AFIRST2, afirst2] > nil
[add2, AADD1, aadd2] > [terms1, s, aterms1]
0 > nil

Status:
aterms1: [1]
first2: [1,2]
AADD1: [1]
terms1: [1]
afirst2: [1,2]
add2: [2,1]
AFIRST2: [1,2]
0: multiset
s: []
aadd2: [2,1]
nil: multiset


The following usable rules [14] were oriented:

a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__sqr(0) → 0
a__add(X1, X2) → add(X1, X2)
mark(s(X)) → s(X)
a__add(0, X) → mark(X)
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__first(0, X) → nil
a__dbl(X) → dbl(X)
mark(0) → 0
mark(recip(X)) → recip(mark(X))
a__sqr(X) → sqr(X)
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__first(X1, X2) → first(X1, X2)
mark(dbl(X)) → a__dbl(mark(X))
a__terms(X) → terms(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__dbl(s(X)) → s(s(dbl(X)))
mark(terms(X)) → a__terms(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(nil) → nil
a__dbl(0) → 0
mark(sqr(X)) → a__sqr(mark(X))
a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__add(s(X), Y) → s(add(X, Y))



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

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

MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(dbl(X)) → MARK(X)
A__TERMS(N) → MARK(N)
MARK(sqr(X)) → MARK(X)
MARK(recip(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof
QDP
                      ↳ QDPOrderProof

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

MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
MARK(recip(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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 oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.

MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
MARK(recip(X)) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
dbl(x1)  =  x1
sqr(x1)  =  x1
recip(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:
cons2 > MARK1

Status:
MARK1: [1]
cons2: [2,1]


The following usable rules [14] were oriented: none



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

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

MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
MARK(recip(X)) → MARK(X)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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 oriented strictly and are deleted.


MARK(recip(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
dbl(x1)  =  x1
sqr(x1)  =  x1
recip(x1)  =  recip(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
[MARK1, recip1]

Status:
recip1: [1]
MARK1: [1]


The following usable rules [14] were oriented: none



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

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

MARK(dbl(X)) → MARK(X)
MARK(sqr(X)) → MARK(X)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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 oriented strictly and are deleted.


MARK(dbl(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(sqr(X)) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
dbl(x1)  =  dbl(x1)
sqr(x1)  =  x1

Lexicographic path order with status [19].
Quasi-Precedence:
[MARK1, dbl1]

Status:
dbl1: [1]
MARK1: [1]


The following usable rules [14] were oriented: none



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

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

MARK(sqr(X)) → MARK(X)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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 oriented strictly and are deleted.


MARK(sqr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
sqr(x1)  =  sqr(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
[MARK1, sqr1]

Status:
sqr1: [1]
MARK1: [1]


The following usable rules [14] were oriented: none



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

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

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(add(sqr(X), dbl(X)))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(X)
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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.