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

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

R
DPs
→DP Problem 1
Polynomial Ordering

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(a) =  1 POL(f(x1, x2)) =  0 POL(F(x1, x2)) =  x2

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Narrowing Transformation

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

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

F(f(a, x), a) -> F(f(f(a, a), f(x, a)), a)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
...
→DP Problem 3
Polynomial Ordering

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(a) =  0 POL(f(x1, x2)) =  1 + x2 POL(F(x1, x2)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
...
→DP Problem 4
Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

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

Rule:

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

Strategy:

innermost

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