Termination w.r.t. Q proof of /tmp/tmpXiceiT/2.26.trs

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:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

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:

F(s(s(x))) → H(g(x))
G(s(x)) → Q(g(x))
G(s(x)) → H(g(x))
F(s(s(x))) → P(g(x))
H(x) → +¹(p(x), q(x))
F(s(s(x))) → Q(g(x))
G(s(x)) → +¹(p(g(x)), q(g(x)))
+¹(x, s(y)) → +¹(x, y)
F(s(s(x))) → P(h(g(x)))
F(s(s(x))) → G(x)
G(s(x)) → P(g(x))
G(s(x)) → G(x)
F(s(s(x))) → +¹(p(g(x)), q(g(x)))
H(x) → P(x)
H(x) → Q(x)

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(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:

F(s(s(x))) → H(g(x))
G(s(x)) → Q(g(x))
G(s(x)) → H(g(x))
F(s(s(x))) → P(g(x))
H(x) → +¹(p(x), q(x))
F(s(s(x))) → Q(g(x))
G(s(x)) → +¹(p(g(x)), q(g(x)))
+¹(x, s(y)) → +¹(x, y)
F(s(s(x))) → P(h(g(x)))
F(s(s(x))) → G(x)
G(s(x)) → P(g(x))
G(s(x)) → G(x)
F(s(s(x))) → +¹(p(g(x)), q(g(x)))
H(x) → P(x)
H(x) → Q(x)

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 13 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

          ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

          ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))
g(0)
g(s(x0))
h(x0)
p(pair(x0, x1))
q(pair(x0, x1))
+(x0, 0)
+(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

          ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

R is empty.
The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))
g(0)
g(s(x0))
h(x0)
p(pair(x0, x1))
q(pair(x0, x1))
+(x0, 0)
+(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(0)
f(s(0))
f(s(s(x0)))
g(0)
g(s(x0))
h(x0)
p(pair(x0, x1))
q(pair(x0, x1))
+(x0, 0)
+(x0, s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

          ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)

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:

+¹(x, s(y)) → +¹(x, y)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))
g(0)
g(s(x0))
h(x0)
p(pair(x0, x1))
q(pair(x0, x1))
+(x0, 0)
+(x0, s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

G(s(x)) → G(x)

R is empty.
The set Q consists of the following terms:

f(0)
f(s(0))
f(s(s(x0)))
g(0)
g(s(x0))
h(x0)
p(pair(x0, x1))
q(pair(x0, x1))
+(x0, 0)
+(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(0)
f(s(0))
f(s(s(x0)))
g(0)
g(s(x0))
h(x0)
p(pair(x0, x1))
q(pair(x0, x1))
+(x0, 0)
+(x0, s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

G(s(x)) → G(x)

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

G(s(x)) → G(x)
The graph contains the following edges 1 > 1