Termination Proof using AProVE

Term Rewriting System R:
[x]
f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(a, x) -> F(b, f(c, x))
F(a, x) -> F(c, x)
F(a, f(b, x)) -> F(b, f(a, x))
F(a, f(b, x)) -> F(a, x)
F(d, f(c, x)) -> F(d, f(a, x))
F(d, f(c, x)) -> F(a, x)
F(a, f(c, x)) -> F(c, f(a, x))
F(a, f(c, x)) -> F(a, x)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

F(a, f(c, x)) -> F(a, x)
F(a, f(b, x)) -> F(a, x)

Rules:

f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(a, f(b, x)) -> F(a, x)

two new Dependency Pairs are created:

F(a, f(b, f(b, x''))) -> F(a, f(b, x''))
F(a, f(b, f(c, x''))) -> F(a, f(c, x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

F(a, f(b, f(c, x''))) -> F(a, f(c, x''))
F(a, f(b, f(b, x''))) -> F(a, f(b, x''))
F(a, f(c, x)) -> F(a, x)

Rules:

f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(a, f(c, x)) -> F(a, x)

three new Dependency Pairs are created:

F(a, f(c, f(c, x''))) -> F(a, f(c, x''))
F(a, f(c, f(b, f(b, x'''')))) -> F(a, f(b, f(b, x'''')))
F(a, f(c, f(b, f(c, x'''')))) -> F(a, f(b, f(c, x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
F(a, f(c, f(b, f(c, x'''')))) -> F(a, f(b, f(c, x'''')))
F(a, f(b, f(b, x''))) -> F(a, f(b, x''))
F(a, f(c, f(b, f(b, x'''')))) -> F(a, f(b, f(b, x'''')))
F(a, f(c, f(c, x''))) -> F(a, f(c, x''))
F(a, f(b, f(c, x''))) -> F(a, f(c, x''))

Rules:

f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Strategy:
innermost
Dependency Pair:
F(d, f(c, x)) -> F(d, f(a, x))

Rules:

f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
F(a, f(c, f(b, f(c, x'''')))) -> F(a, f(b, f(c, x'''')))
F(a, f(b, f(b, x''))) -> F(a, f(b, x''))
F(a, f(c, f(b, f(b, x'''')))) -> F(a, f(b, f(b, x'''')))
F(a, f(c, f(c, x''))) -> F(a, f(c, x''))
F(a, f(b, f(c, x''))) -> F(a, f(c, x''))

Rules:

f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Strategy:
innermost
Dependency Pair:
F(d, f(c, x)) -> F(d, f(a, x))

Rules:

f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes