Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(x'''', s(s(s(y'''')))) -> +'(x'''', s(s(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)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

Dependency Pair:

Rules:

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

         ↳Rw

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(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 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Rw

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(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 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Rw

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(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 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 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Rw

Dependency Pair:

Rules:

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

         ↳Rewriting Transformation

Dependency Pair:

FACT(s(x)) -> FACT(p(s(x)))

Rules:

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

Strategy:

innermost

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

FACT(s(x)) -> FACT(p(s(x)))

one new Dependency Pair is created:

FACT(s(x)) -> FACT(x)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

           →DP Problem 10

             ↳Forward Instantiation Transformation

Dependency Pair:

FACT(s(x)) -> FACT(x)

Rules:

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

Strategy:

innermost

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

FACT(s(x)) -> FACT(x)

one new Dependency Pair is created:

FACT(s(s(x''))) -> FACT(s(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Rw

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 11

                 ↳Polynomial Ordering

Dependency Pair:

FACT(s(s(x''))) -> FACT(s(x''))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

FACT(s(s(x''))) -> FACT(s(x''))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FACT(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 3

         ↳Rw

           →DP Problem 10

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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