Termination w.r.t. Q proof of TRS/TRCSREx26_Luc03b

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

Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
dbl(x1) = x1
first(x1, x2) = first(x1, x2)
A__FIRST(x1, x2) = A__FIRST(x2)
mark(x1) = x1
sqr(x1) = x1
add(x1, x2) = add(x1, x2)
recip(x1) = x1
terms(x1) = x1
s(x1) = s
cons(x1, x2) = x1
A__ADD(x1, x2) = A__ADD(x2)
0 = 0
A__TERMS(x1) = A__TERMS(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) = x1

Lexicographic Path Order [19].
Precedence:

[MARK₁, first₂, A_{_FIRST₁}, add₂, s, A_{_ADD₁}, 0, A_{_TERMS₁}, a_{_add₂}, a_{_first₂}, nil]

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

A__TERMS(N) → MARK(N)
MARK(terms(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

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

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

Lexicographic Path Order [19].
Precedence:

[A_{_TERMS₁}, terms₁, a_{_{terms_1]}} > s
0 > s
[a_{_first₂}, nil, first_2] > s

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

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

MARK(dbl(X)) → MARK(X)
MARK(terms(X)) → A__TERMS(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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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 [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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 [19].
Precedence:

[MARK₁, recip_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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 [19].
Precedence:

[MARK₁, dbl_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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 [19].
Precedence:

[MARK₁, sqr_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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.