Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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)

one new Dependency Pair is created:

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

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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 AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = 1 + 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

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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 Pair:

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

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(-'(x₁, x₂)) = 1 + 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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, 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 Pair:

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

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Strategy:

innermost

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

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

one new Dependency Pair is created:

EXP(x'', s(s(y''))) -> EXP(x'', s(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 10

             ↳Forward Instantiation Transformation

Dependency Pair:

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

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Strategy:

innermost

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

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

one new Dependency Pair is created:

EXP(x'''', s(s(s(y'''')))) -> EXP(x'''', s(s(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 10

             ↳FwdInst

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

Dependency Pair:

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

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(EXP(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

exp(x, 0) -> s(0)
exp(x, s(y)) -> *(x, exp(x, y))
*(0, y) -> 0
*(s(x), y) -> +(y, *(x, y))
-(0, y) -> 0
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

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

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

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