Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

Rule:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

Rule:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

Rule:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

F(a, f(f(x'', a), a)) -> F(a, f(x'', a))
F(a, f(f(f(f(x''', a), a), a), a)) -> F(a, f(f(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 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

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

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

Rule:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

Additionally, the following usable rule for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(a) = 0

POL(f(x₁, x₂)) = 1 + x₁

POL(F(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rule:

f(a, f(x, a)) -> f(a, f(f(a, x), f(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 + x₁
POL(F(x₁, x₂))	= x₂