Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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:

:¹(z, +(x, f(y))) → :¹(g(z, y), +(x, a))
:¹(+(x, y), z) → :¹(y, z)
:¹(:(x, y), z) → :¹(y, z)
:¹(:(x, y), z) → :¹(x, :(y, z))
:¹(+(x, y), z) → :¹(x, z)

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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:

:¹(z, +(x, f(y))) → :¹(g(z, y), +(x, a))
:¹(+(x, y), z) → :¹(y, z)
:¹(:(x, y), z) → :¹(y, z)
:¹(:(x, y), z) → :¹(x, :(y, z))
:¹(+(x, y), z) → :¹(x, z)

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPSizeChangeProof

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

:¹(+(x, y), z) → :¹(y, z)
:¹(:(x, y), z) → :¹(y, z)
:¹(:(x, y), z) → :¹(x, :(y, z))
:¹(+(x, y), z) → :¹(x, z)

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

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