Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(s(x), y) -> +'(x, y)
+'(p(x), y) -> +'(x, y)
MINUS(s(x)) -> MINUS(x)
MINUS(p(x)) -> MINUS(x)
*'(s(x), y) -> +'(*(x, y), y)
*'(s(x), y) -> *'(x, y)
*'(p(x), y) -> +'(*(x, y), minus(y))
*'(p(x), y) -> *'(x, y)
*'(p(x), y) -> MINUS(y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 10

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = x₁

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 12

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(p(x)) -> MINUS(x)
MINUS(s(x)) -> MINUS(x)

Rules:

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

Strategy:

innermost

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

MINUS(s(x)) -> MINUS(x)

two new Dependency Pairs are created:

MINUS(s(s(x''))) -> MINUS(s(x''))
MINUS(s(p(x''))) -> MINUS(p(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(s(p(x''))) -> MINUS(p(x''))
MINUS(s(s(x''))) -> MINUS(s(x''))
MINUS(p(x)) -> MINUS(x)

Rules:

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

Strategy:

innermost

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

MINUS(p(x)) -> MINUS(x)

three new Dependency Pairs are created:

MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(x''))) -> MINUS(s(x''))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(s(p(x''))) -> MINUS(p(x''))

Rules:

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

Strategy:

innermost

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

MINUS(s(s(x''))) -> MINUS(s(x''))

two new Dependency Pairs are created:

MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(s(p(x''))) -> MINUS(p(x''))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))

Rules:

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

Strategy:

innermost

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

MINUS(s(p(x''))) -> MINUS(p(x''))

three new Dependency Pairs are created:

MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(x''))) -> MINUS(p(x''))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))

Rules:

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

Strategy:

innermost

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

MINUS(p(p(x''))) -> MINUS(p(x''))

three new Dependency Pairs are created:

MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

Rules:

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

Strategy:

innermost

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

MINUS(p(s(s(x'''')))) -> MINUS(s(s(x'''')))

two new Dependency Pairs are created:

MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 19

                 ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

Rules:

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

Strategy:

innermost

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

MINUS(p(s(p(x'''')))) -> MINUS(s(p(x'''')))

three new Dependency Pairs are created:

MINUS(p(s(p(p(x''''''))))) -> MINUS(s(p(p(x''''''))))
MINUS(p(s(p(s(s(x'''''''')))))) -> MINUS(s(p(s(s(x'''''''')))))
MINUS(p(s(p(s(p(x'''''''')))))) -> MINUS(s(p(s(p(x'''''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 20

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(p(s(p(s(p(x'''''''')))))) -> MINUS(s(p(s(p(x'''''''')))))
MINUS(p(s(p(s(s(x'''''''')))))) -> MINUS(s(p(s(s(x'''''''')))))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))
MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(p(x''''''))))) -> MINUS(s(p(p(x''''''))))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MINUS(p(s(p(s(p(x'''''''')))))) -> MINUS(s(p(s(p(x'''''''')))))
MINUS(p(s(p(s(s(x'''''''')))))) -> MINUS(s(p(s(s(x'''''''')))))
MINUS(p(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))
MINUS(p(s(s(p(x''''''))))) -> MINUS(s(s(p(x''''''))))
MINUS(p(s(s(s(x''''''))))) -> MINUS(s(s(s(x''''''))))
MINUS(p(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(p(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(p(s(p(p(x''''''))))) -> MINUS(s(p(p(x''''''))))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁)) = 1 + x₁

POL(s(x₁)) = x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 21

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

Dependency Pairs:

MINUS(s(p(s(s(x''''''))))) -> MINUS(p(s(s(x''''''))))
MINUS(s(s(p(x'''')))) -> MINUS(s(p(x'''')))
MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))
MINUS(s(p(p(x'''')))) -> MINUS(p(p(x'''')))
MINUS(s(p(s(p(x''''''))))) -> MINUS(p(s(p(x''''''))))

Rules:

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

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 22

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(s(s(x'''')))) -> MINUS(s(s(x'''')))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 23

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 25

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 26

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 27

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 28

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 29

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 30

                 ↳Polynomial Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 31

                 ↳Dependency Graph

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 32

                 ↳Polynomial Ordering

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 24

             ↳FwdInst

...

               →DP Problem 33

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes

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

POL(MINUS(x₁))	= 1 + x₁
POL(s(x₁))	= x₁
POL(p(x₁))	= 1 + x₁

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