Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

Rules:

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

Strategy:
innermost
Dependency Pairs:
F(b, f(a, x)) -> F(b, x)
F(b, f(a, x)) -> F(b, f(b, x))
F(b, f(a, x)) -> F(b, f(b, f(b, x)))

Rules:

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

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

Rules:

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

Strategy:
innermost
Dependency Pairs:
F(b, f(a, x)) -> F(b, x)
F(b, f(a, x)) -> F(b, f(b, x))
F(b, f(a, x)) -> F(b, f(b, f(b, x)))

Rules:

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

Strategy:
innermost

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