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

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

R
DPs
→DP Problem 1
Narrowing Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

A(x, g) -> A(f, a(g, a(f, x)))
two new Dependency Pairs are created:

A(a(f, x''), g) -> A(f, a(g, a(x'', g)))
A(g, g) -> A(f, a(g, a(f, a(g, a(f, f)))))

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
Nar
→DP Problem 2
Usable Rules (Innermost)

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

R
DPs
→DP Problem 1
Nar
→DP Problem 2
UsableRules
...
→DP Problem 3
Modular Removal of Rules

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(g) =  0 POL(a(x1, x2)) =  x1 + x2 POL(A(x1, x2)) =  x1 + x2 POL(f) =  0

We have the following set D of usable symbols: {g, A, f}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

No Rules can be deleted.

The result of this processor delivers one new DP problem.

R
DPs
→DP Problem 1
Nar
→DP Problem 2
UsableRules
...
→DP Problem 4
Modular Removal of Rules

Dependency Pair:

A(x, g) -> A(f, x)

Rule:

none

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(g) =  0 POL(A(x1, x2)) =  x1 + x2 POL(f) =  0

We have the following set D of usable symbols: {A, f}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

A(x, g) -> A(f, x)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes