Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Instantiation Transformation

       →DP Problem 2

         ↳Inst

Dependency Pair:

F(s(x), y, z) -> F(x, s(c(y)), c(z))

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

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

F(s(x), y, z) -> F(x, s(c(y)), c(z))

one new Dependency Pair is created:

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 3

             ↳Instantiation Transformation

       →DP Problem 2

         ↳Inst

Dependency Pair:

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

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

F(s(x''), s(c(y'')), c(z'')) -> F(x'', s(c(s(c(y'')))), c(c(z'')))

one new Dependency Pair is created:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 3

             ↳Inst

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Inst

Dependency Pair:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(s(x''''), s(c(s(c(y'''')))), c(c(z''''))) -> F(x'''', s(c(s(c(s(c(y'''')))))), c(c(c(z''''))))

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(c(x₁)) = 0

POL(s(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 3

             ↳Inst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Inst

Dependency Pair:

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Instantiation Transformation

Dependency Pair:

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

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

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

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

one new Dependency Pair is created:

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''), y'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Inst

           →DP Problem 6

             ↳Forward Instantiation Transformation

Dependency Pair:

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''), y'')

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

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

F(c(y''), c(c(y'')), c(y'')) -> F(y'', c(y''), y'')

one new Dependency Pair is created:

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(y'''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Inst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pair:

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(y'''''))

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(c(c(y''''')), c(c(c(y'''''))), c(c(y'''''))) -> F(c(y'''''), c(c(y''''')), c(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(c(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Inst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
f(s(x), y, z) -> f(x, s(c(y)), c(z))
f(c(x), x, y) -> c(y)
g(x, y) -> x
g(x, y) -> y

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(c(x₁))	= 0
POL(s(x₁))	= 1 + x₁
POL(F(x₁, x₂, x₃))	= x₁