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 w.r.t. to the AFS 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 w.r.t. to the AFS 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 w.r.t. to the AFS 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

                 ↳Remaining Obligation(s)

The following remains to be proven:
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

Innermost Termination of R could not be shown.
Duration:
0:01 minutes

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

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