Termination Proof using AProVE

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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

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

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:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

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(f(a, a), f(x, 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

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(a, a), f(x, a))

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

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

one new Dependency Pair is created:

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

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rule:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(a)	= 0
POL(f(x₁, x₂))	= 1 + x₂
POL(F(x₁, x₂))	= x₁