Termination Proof using AProVE

Term Rewriting System R:
[x, y]
p(s(x)) -> x
s(p(x)) -> 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) -> S(+(x, y))
+'(s(x), y) -> +'(x, y)
+'(p(x), y) -> P(+(x, y))
+'(p(x), y) -> +'(x, y)
MINUS(s(x)) -> P(minus(x))
MINUS(s(x)) -> MINUS(x)
MINUS(p(x)) -> S(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 six SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

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)

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 7

             ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

p(s(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(p(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 7

             ↳Polo

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

Rules:

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

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

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

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)

one new Dependency Pair is created:

MINUS(p(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 9

             ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

p(s(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳Polo

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

Rules:

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

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

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

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 3

         ↳FwdInst

           →DP Problem 11

             ↳Polynomial Ordering

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

s(p(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(p(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳Polo

...

               →DP Problem 12

                 ↳Dependency Graph

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

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)

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 13

             ↳Polynomial Ordering

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

s(p(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁)) = x₁

POL(s(x₁)) = 1 + 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 4

         ↳FwdInst

           →DP Problem 13

             ↳Polo

...

               →DP Problem 14

                 ↳Dependency Graph

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Forward Instantiation Transformation

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

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)

one new Dependency Pair is created:

*'(p(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 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

           →DP Problem 15

             ↳Polynomial Ordering

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

p(s(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + 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 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

           →DP Problem 15

             ↳Polo

...

               →DP Problem 16

                 ↳Dependency Graph

       →DP Problem 6

         ↳FwdInst

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳Forward Instantiation Transformation

Dependency Pair:

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

Rules:

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

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 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

           →DP Problem 17

             ↳Polynomial Ordering

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

s(p(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + 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 4

         ↳FwdInst

       →DP Problem 5

         ↳FwdInst

       →DP Problem 6

         ↳FwdInst

           →DP Problem 17

             ↳Polo

...

               →DP Problem 18

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

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

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

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