Termination w.r.t. Q proof of /tmp/tmpQ931uk/2.27.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:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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:

SP(pair(x, y)) → +¹(x, y)
FIB(s(s(x))) → G(x)
G(s(x)) → G(x)
+¹(x, s(y)) → +¹(x, y)
G(s(x)) → NP(g(x))
NP(pair(x, y)) → +¹(x, y)
FIB(s(s(x))) → SP(g(x))

The TRS R consists of the following rules:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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:

SP(pair(x, y)) → +¹(x, y)
FIB(s(s(x))) → G(x)
G(s(x)) → G(x)
+¹(x, s(y)) → +¹(x, y)
G(s(x)) → NP(g(x))
NP(pair(x, y)) → +¹(x, y)
FIB(s(s(x))) → SP(g(x))

The TRS R consists of the following rules:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ 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

            ↳ 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:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ 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