Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(f(a, b), x) -> F(a, f(a, x))
F(f(a, b), x) -> F(a, x)
F(f(b, a), x) -> F(b, f(b, x))
F(f(b, a), x) -> F(b, x)
F(x, f(y, z)) -> F(f(x, y), z)
F(x, f(y, z)) -> F(x, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(b, a), x) -> F(b, x)
F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, x)
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), x) -> F(a, f(a, x))

Rules:

f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

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

one new Dependency Pair is created:

F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(b, a), x) -> F(b, x)

Rules:

f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

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

one new Dependency Pair is created:

F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))
F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))

Rules:

f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

F(x, f(y, z)) -> F(x, y)

five new Dependency Pairs are created:

F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(b, a), f(y', z)) -> F(f(b, a), y')
F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))
F(f(b, a), f(f(y'''', z''''), z)) -> F(f(b, a), f(y'''', z''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pairs:

F(f(b, a), f(f(y'''', z''''), z)) -> F(f(b, a), f(y'''', z''''))
F(f(b, a), f(y', z)) -> F(f(b, a), y')
F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
F(f(b, a), x) -> F(b, f(b, x))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))

Rules:

f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(f(b, a), f(f(y'''', z''''), z)) -> F(f(b, a), f(y'''', z''''))
F(f(b, a), f(y', z)) -> F(f(b, a), y')
F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))

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

f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(b) = 0

POL(a) = 0

POL(f(x₁, x₂)) = 1 + x₁ + x₂

POL(F(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Instantiation Transformation

Dependency Pairs:

F(f(b, a), x) -> F(b, f(b, x))
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)

Rules:

f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

F(x, f(y, z)) -> F(f(x, y), z)

two new Dependency Pairs are created:

F(a, f(y', z')) -> F(f(a, y'), z')
F(b, f(y', z')) -> F(f(b, y'), z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(a, f(y', z')) -> F(f(a, y'), z')
F(f(a, b), x) -> F(a, f(a, x))
F(b, f(y', z')) -> F(f(b, y'), z')
F(f(b, a), x) -> F(b, f(b, x))

Rules:

f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

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