Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
p(s(x)) -> x
f(s(x)) -> f(-(p(*(s(x), s(x))), *(s(x), s(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: Polynomial ordering with Polynomial interpretation:

POL(-'(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = x₁ + x₂

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 2

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳Rw

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rewriting Transformation

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

F(s(x)) -> F(-(p(+(s(x), *(s(x), x))), *(s(x), s(x))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rewriting Transformation

Dependency Pair:

F(s(x)) -> F(-(p(+(s(x), *(s(x), x))), *(s(x), s(x))))

Rules:

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

Strategy:

innermost

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

F(s(x)) -> F(-(p(+(s(x), *(s(x), x))), *(s(x), s(x))))

one new Dependency Pair is created:

F(s(x)) -> F(-(p(s(+(x, *(s(x), x)))), *(s(x), s(x))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 9

                 ↳Rewriting Transformation

Dependency Pair:

F(s(x)) -> F(-(p(s(+(x, *(s(x), x)))), *(s(x), s(x))))

Rules:

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

Strategy:

innermost

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

F(s(x)) -> F(-(p(s(+(x, *(s(x), x)))), *(s(x), s(x))))

one new Dependency Pair is created:

F(s(x)) -> F(-(+(x, *(s(x), x)), *(s(x), s(x))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 10

                 ↳Rewriting Transformation

Dependency Pair:

F(s(x)) -> F(-(+(x, *(s(x), x)), *(s(x), s(x))))

Rules:

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

Strategy:

innermost

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

F(s(x)) -> F(-(+(x, *(s(x), x)), *(s(x), s(x))))

one new Dependency Pair is created:

F(s(x)) -> F(-(+(x, *(s(x), x)), +(s(x), *(s(x), x))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 11

                 ↳Rewriting Transformation

Dependency Pair:

F(s(x)) -> F(-(+(x, *(s(x), x)), +(s(x), *(s(x), x))))

Rules:

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

Strategy:

innermost

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

F(s(x)) -> F(-(+(x, *(s(x), x)), +(s(x), *(s(x), x))))

one new Dependency Pair is created:

F(s(x)) -> F(-(+(x, *(s(x), x)), s(+(x, *(s(x), x)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pair:

F(s(x)) -> F(-(+(x, *(s(x), x)), s(+(x, *(s(x), x)))))

Rules:

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

Strategy:

innermost

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

F(s(x)) -> F(-(+(x, *(s(x), x)), s(+(x, *(s(x), x)))))

eight new Dependency Pairs are created:

F(s(0)) -> F(-(*(s(0), 0), s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(+(s(x''), *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(+(0, 0), s(+(0, *(s(0), 0)))))
F(s(s(y'))) -> F(-(+(s(y'), +(s(s(y')), *(s(s(y')), y'))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(*(s(0), 0))))
F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 13

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(s(y'), +(s(s(y')), *(s(s(y')), y'))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(+(0, 0), s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(+(s(x''), *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, *(s(0), 0)))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(*(s(0), 0), s(+(0, *(s(0), 0)))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 14

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(s(y'), +(s(s(y')), *(s(s(y')), y'))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(+(0, 0), s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(+(s(x''), *(s(s(x'')), s(x''))))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(+(s(x''), *(s(s(x'')), s(x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 15

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(s(y'), +(s(s(y')), *(s(s(y')), y'))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(+(0, 0), s(+(0, *(s(0), 0)))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(+(0, 0), s(+(0, *(s(0), 0)))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 16

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(s(y'), +(s(s(y')), *(s(s(y')), y'))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(s(y'), +(s(s(y')), *(s(s(y')), y'))), s(+(s(y'), *(s(s(y')), s(y'))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 17

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(+(0, *(s(0), 0)), s(*(s(0), 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 18

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(s(x''), *(s(s(x'')), s(x''))), s(s(+(x'', *(s(s(x'')), s(x'')))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 19

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))
F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(+(0, *(s(0), 0)), s(+(0, 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 20

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))
F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(s(y'), *(s(s(y')), s(y'))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 21

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(*(s(0), 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 22

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), +(s(x''), *(s(s(x'')), s(x'')))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 23

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(+(0, *(s(0), 0)))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(*(s(0), 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 24

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(s(+(y', +(s(s(y')), *(s(s(y')), y')))), s(+(s(y'), *(s(s(y')), s(y'))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 25

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(*(s(0), 0), s(*(s(0), 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(*(s(0), 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 26

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))
F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(s(+(x'', *(s(s(x'')), s(x'')))), s(s(+(x'', *(s(s(x'')), s(x'')))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 27

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(*(s(0), 0), s(+(0, 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(+(0, 0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 28

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(+(0, 0))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(s(+(y', *(s(s(y')), s(y')))), s(+(s(y'), +(s(s(y')), *(s(s(y')), y'))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 29

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(+(0, 0))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(*(s(0), 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 30

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(+(0, 0))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), +(s(x''), *(s(s(x'')), s(x'')))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 31

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(+(0, 0))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(0)))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(*(s(0), 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 32

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(+(0, 0))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), *(s(s(y')), s(y')))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 33

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(+(0, 0))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(*(s(0), 0))))
F(s(0)) -> F(-(0, s(0)))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(*(s(0), 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 34

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(+(0, 0))))
F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', *(s(s(x'')), s(x''))), s(+(x'', *(s(s(x'')), s(x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), s(+(x'', *(s(s(x'')), s(x''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 35

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(+(0, 0))))
F(s(0)) -> F(-(0, s(0)))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(+(0, 0))))

one new Dependency Pair is created:

F(s(0)) -> F(-(0, s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 36

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), s(+(x'', *(s(s(x'')), s(x''))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', *(s(s(y')), s(y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 37

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), +(s(x''), *(s(s(x'')), s(x'')))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), +(s(x''), *(s(s(x'')), s(x'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 38

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), *(s(s(y')), s(y')))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), *(s(s(y')), s(y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 39

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), *(s(s(y')), s(y')))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), +(s(x''), *(s(s(x'')), s(x'')))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', +(s(s(x'')), *(s(s(x'')), x''))), s(+(x'', *(s(s(x'')), s(x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), s(+(x'', *(s(s(x'')), s(x''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 40

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), *(s(s(y')), s(y')))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', +(s(s(y')), *(s(s(y')), y'))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 41

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), *(s(s(y')), s(y')))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), +(s(x''), *(s(s(x'')), s(x'')))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), s(+(x'', *(s(s(x'')), s(x''))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), +(s(x''), *(s(s(x'')), s(x'')))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 42

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), *(s(s(y')), s(y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), *(s(s(y')), s(y')))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', *(s(s(y')), s(y'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 43

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', *(s(s(y')), s(y'))))))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(+(s(x''), *(s(s(x'')), x'')))), s(+(x'', *(s(s(x'')), s(x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 44

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', *(s(s(y')), s(y'))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(+(s(y'), *(s(s(y')), y')))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 45

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', *(s(s(y')), s(y'))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 46

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', *(s(s(y')), s(y'))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', *(s(s(y')), s(y'))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 47

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', *(s(s(x'')), s(x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 48

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), +(s(y'), +(s(s(y')), *(s(s(y')), y')))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 49

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 50

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 51

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', +(s(s(x'')), *(s(s(x'')), x''))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 52

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', +(s(s(y')), *(s(s(y')), y'))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 53

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 54

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 55

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))

Rules:

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

Strategy:

innermost

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

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(+(s(x''), *(s(s(x'')), x'')))))))

one new Dependency Pair is created:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 56

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))

Rules:

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

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(+(s(y'), *(s(s(y')), y')))))))

one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 57

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(0)))

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 58

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(0)))

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 59

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(0)) -> F(-(0, s(0)))
F(s(0)) -> F(-(0, s(0)))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(0)))

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 60

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(0)) -> F(-(0, s(0)))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))

Rules:

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

Strategy:

innermost

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

F(s(0)) -> F(-(0, s(0)))

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Rw

           →DP Problem 8

             ↳Rw

...

               →DP Problem 61

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(y'))) -> F(-(+(y', s(s(+(y', *(s(s(y')), y'))))), s(+(y', s(s(+(y', *(s(s(y')), y'))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))
F(s(s(x''))) -> F(-(+(x'', s(s(+(x'', *(s(s(x'')), x''))))), s(+(x'', s(s(+(x'', *(s(s(x'')), x''))))))))

Rules:

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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:19 minutes

POL(-'(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(s(x₁))	= 1 + x₁
POL(+'(x₁, x₂))	= x₁ + x₂