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))

Termination of R to be shown.

R
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.

R
DPs
→DP Problem 1
Polynomial Ordering
→DP Problem 2
Nar

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))

The following dependency pairs can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 3
Dependency Graph
→DP Problem 2
Nar

Dependency Pair:

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))

Using the Dependency Graph resulted in no new DP problems.

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

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))

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

F(d, f(c, x)) -> F(d, f(a, x))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
→DP Problem 4
Narrowing Transformation

Dependency Pairs:

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

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

F(d, f(c, x'')) -> F(d, f(b, f(c, x'')))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
→DP Problem 4
Nar
...
→DP Problem 5
Polynomial Ordering

Dependency Pairs:

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

The following dependency pair can be strictly oriented:

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

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

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))

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

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
→DP Problem 4
Nar
...
→DP Problem 6
Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

F(d, f(c, f(c, x''))) -> F(d, f(c, 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))

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