Termination Proof using AProVE

Term Rewriting System R:
[x, y]
f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(f(0, x), 1) -> F(g(f(x, x)), x)
F(f(0, x), 1) -> F(x, x)
F(g(x), y) -> F(x, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(0, x), 1) -> F(x, x)
F(g(x), y) -> F(x, y)
F(f(0, x), 1) -> F(g(f(x, x)), x)

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

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

F(f(0, x), 1) -> F(x, x)

one new Dependency Pair is created:

F(f(0, g(x'')), 1) -> F(g(x''), g(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(0, g(x'')), 1) -> F(g(x''), g(x''))
F(f(0, x), 1) -> F(g(f(x, x)), x)
F(g(x), y) -> F(x, y)

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

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

F(g(x), y) -> F(x, y)

three new Dependency Pairs are created:

F(g(g(x'')), y'') -> F(g(x''), y'')
F(g(f(0, x'')), 1) -> F(f(0, x''), 1)
F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)
F(f(0, x), 1) -> F(g(f(x, x)), x)
F(g(f(0, x'')), 1) -> F(f(0, x''), 1)
F(g(g(x'')), y'') -> F(g(x''), y'')
F(f(0, g(x'')), 1) -> F(g(x''), g(x''))

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

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

F(f(0, x), 1) -> F(g(f(x, x)), x)

one new Dependency Pair is created:

F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))
F(g(f(0, x'')), 1) -> F(f(0, x''), 1)
F(g(g(x'')), y'') -> F(g(x''), y'')
F(f(0, g(x'')), 1) -> F(g(x''), g(x''))
F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

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

F(f(0, g(x'')), 1) -> F(g(x''), g(x''))

one new Dependency Pair is created:

F(f(0, g(g(x''''))), 1) -> F(g(g(x'''')), g(g(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)
F(f(0, g(g(x''''))), 1) -> F(g(g(x'''')), g(g(x'''')))
F(g(f(0, x'')), 1) -> F(f(0, x''), 1)
F(g(g(x'')), y'') -> F(g(x''), y'')
F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

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

F(g(g(x'')), y'') -> F(g(x''), y'')

three new Dependency Pairs are created:

F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')
F(g(g(f(0, x''''))), 1) -> F(g(f(0, x'''')), 1)
F(g(g(f(0, g(x'''''')))), 1) -> F(g(f(0, g(x''''''))), 1)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Narrowing Transformation

Dependency Pairs:

F(g(g(f(0, g(x'''''')))), 1) -> F(g(f(0, g(x''''''))), 1)
F(f(0, g(g(x''''))), 1) -> F(g(g(x'''')), g(g(x'''')))
F(g(f(0, x'')), 1) -> F(f(0, x''), 1)
F(g(g(f(0, x''''))), 1) -> F(g(f(0, x'''')), 1)
F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')
F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))
F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

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

F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))

one new Dependency Pair is created:

F(f(0, g(g(x'))), 1) -> F(g(g(g(f(x', g(g(x')))))), g(g(x')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

F(f(0, g(g(x'))), 1) -> F(g(g(g(f(x', g(g(x')))))), g(g(x')))
F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)
F(g(g(f(0, x''''))), 1) -> F(g(f(0, x'''')), 1)
F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')
F(f(0, g(g(x''''))), 1) -> F(g(g(x'''')), g(g(x'''')))
F(g(f(0, x'')), 1) -> F(f(0, x''), 1)
F(g(g(f(0, g(x'''''')))), 1) -> F(g(f(0, g(x''''''))), 1)

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(f(0, g(g(x'))), 1) -> F(g(g(g(f(x', g(g(x')))))), g(g(x')))
F(f(0, g(g(x''''))), 1) -> F(g(g(x'''')), g(g(x'''')))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(g(x₁)) = 0

POL(1) = 1

POL(f(x₁, x₂)) = 0

POL(F(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pairs:

F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)
F(g(g(f(0, x''''))), 1) -> F(g(f(0, x'''')), 1)
F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')
F(g(f(0, x'')), 1) -> F(f(0, x''), 1)
F(g(g(f(0, g(x'''''')))), 1) -> F(g(f(0, g(x''''''))), 1)

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 9

                 ↳Polynomial Ordering

Dependency Pair:

F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(g(g(g(x''''))), y'''') -> F(g(g(x'''')), y'''')

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(g(x₁)) = 1 + x₁

POL(F(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

Dependency Pair:

Rules:

f(f(0, x), 1) -> f(g(f(x, x)), x)
f(g(x), y) -> g(f(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(0)	= 0
POL(g(x₁))	= 0
POL(1)	= 1
POL(f(x₁, x₂))	= 0
POL(F(x₁, x₂))	= x₂