Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) -> P(-(s(x), y))
F(s(x), y) -> -'(s(x), y)
F(s(x), y) -> P(-(y, s(x)))
F(s(x), y) -> -'(y, s(x))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) -> P(-(x, s(y)))
F(x, s(y)) -> -'(x, s(y))
F(x, s(y)) -> P(-(s(y), x))
F(x, s(y)) -> -'(s(y), x)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(x, y)

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

-'(x₁, x₂) -> -'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))

three new Dependency Pairs are created:

F(s(x''), 0) -> F(p(s(x'')), p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rewriting Transformation

Dependency Pairs:

F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x''), 0) -> F(p(s(x'')), p(-(0, s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x''), 0) -> F(p(s(x'')), p(-(0, s(x''))))

one new Dependency Pair is created:

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 5

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))

one new Dependency Pair is created:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 6

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))

one new Dependency Pair is created:

F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))

three new Dependency Pairs are created:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 8

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))

one new Dependency Pair is created:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 9

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))

one new Dependency Pair is created:

F(0, s(y')) -> F(p(-(0, s(y'))), y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 10

                 ↳Rewriting Transformation

Dependency Pairs:

F(0, s(y')) -> F(p(-(0, s(y'))), y')
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))

one new Dependency Pair is created:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 11

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(0, s(y')) -> F(p(-(0, s(y'))), y')

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pairs:

F(0, s(y')) -> F(p(-(0, s(y'))), y')
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(0, s(y')) -> F(p(-(0, s(y'))), y')

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 13

                 ↳Argument Filtering and Ordering

Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

The following usable rules for innermost can be oriented:

p(s(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

s > p

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂) -> F(x₁, x₂)
s(x₁) -> s(x₁)
p(x₁) -> p(x₁)
-(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Rw

...

               →DP Problem 14

                 ↳Dependency Graph

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes