Termination w.r.t. Q proof of /tmp/tmpnnHBEo/Ex2_Luc02a

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(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(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(mark(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(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(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(mark(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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(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(mark(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

      ↳ QDPOrderProof

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

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

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(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(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(mark(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 [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

MARK(cons(X1, X2)) → MARK(X1)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(terms(X)) → A__TERMS(mark(X))
MARK(recip(X)) → MARK(X)

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

Recursive path order with status [2].
Quasi-Precedence:

0 > nil
[terms₁, A_{_TERMS₁}, a_{_terms₁}] > [sqr₁, A_{_SQR₁}, dbl₁, a_{_sqr₁}, a_{_dbl₁}] > [add₂, a_{_add₂}] > A_{_ADD₂}
[terms₁, A_{_TERMS₁}, a_{_terms₁}] > [sqr₁, A_{_SQR₁}, dbl₁, a_{_sqr₁}, a_{_dbl₁}] > [add₂, a_{_add₂}] > s₁
[first₂, a_{_first₂}]

Status:

A_{_TERMS₁}: multiset
sqr₁: [1]
a_{_dbl₁}: [1]
a_{_add₂}: multiset
first₂: multiset
0: multiset
terms₁: multiset
a_{_first₂}: multiset
add₂: multiset
A_{_ADD₂}: multiset
a_{_terms₁}: multiset
dbl₁: [1]
s₁: multiset
A_{_SQR₁}: [1]
nil: multiset
a_{_sqr₁}: [1]

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(terms(X)) → A__TERMS(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
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(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(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(mark(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 [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

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

MARK(cons(X1, X2)) → MARK(X1)
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(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(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(mark(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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QDPSizeChangeProof

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
MARK(recip(X)) → MARK(X)
The graph contains the following edges 1 > 1