Termination Proof using AProVE

Term Rewriting System R:
[X, Y]
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(X), s(Y)) -> P(minus(X, Y))
MINUS(s(X), s(Y)) -> MINUS(X, Y)
DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(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), s(Y)) -> MINUS(X, Y)

one new Dependency Pair is created:

MINUS(s(s(X'')), s(s(Y''))) -> MINUS(s(X''), s(Y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

MINUS(s(s(X'')), s(s(Y''))) -> MINUS(s(X''), s(Y''))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(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'')), s(s(Y''))) -> MINUS(s(X''), s(Y''))

one new Dependency Pair is created:

MINUS(s(s(s(X''''))), s(s(s(Y'''')))) -> MINUS(s(s(X'''')), s(s(Y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

MINUS(s(s(s(X''''))), s(s(s(Y'''')))) -> MINUS(s(s(X'''')), s(s(Y'''')))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(s(s(X''''))), s(s(s(Y'''')))) -> MINUS(s(s(X'''')), s(s(Y'''')))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

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

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

two new Dependency Pairs are created:

DIV(s(X''), s(0)) -> DIV(X'', s(0))
DIV(s(s(X'')), s(s(Y''))) -> DIV(p(minus(X'', Y'')), s(s(Y'')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Forward Instantiation Transformation

           →DP Problem 7

             ↳Nar

Dependency Pair:

DIV(s(X''), s(0)) -> DIV(X'', s(0))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

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

DIV(s(X''), s(0)) -> DIV(X'', s(0))

one new Dependency Pair is created:

DIV(s(s(X'''')), s(0)) -> DIV(s(X''''), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

           →DP Problem 7

             ↳Nar

Dependency Pair:

DIV(s(s(X'''')), s(0)) -> DIV(s(X''''), s(0))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

DIV(s(s(X'''')), s(0)) -> DIV(s(X''''), s(0))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(DIV(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 11

                 ↳Dependency Graph

           →DP Problem 7

             ↳Nar

Dependency Pair:

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳FwdInst

           →DP Problem 7

             ↳Narrowing Transformation

Dependency Pair:

DIV(s(s(X'')), s(s(Y''))) -> DIV(p(minus(X'', Y'')), s(s(Y'')))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

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

DIV(s(s(X'')), s(s(Y''))) -> DIV(p(minus(X'', Y'')), s(s(Y'')))

two new Dependency Pairs are created:

DIV(s(s(X''')), s(s(0))) -> DIV(p(X'''), s(s(0)))
DIV(s(s(s(X'))), s(s(s(Y')))) -> DIV(p(p(minus(X', Y'))), s(s(s(Y'))))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳FwdInst

           →DP Problem 7

             ↳Nar

...

               →DP Problem 9

                 ↳Polynomial Ordering

Dependency Pair:

DIV(s(s(X''')), s(s(0))) -> DIV(p(X'''), s(s(0)))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

DIV(s(s(X''')), s(s(0))) -> DIV(p(X'''), s(s(0)))

Additionally, the following usable rule for innermost can be oriented:

p(s(X)) -> X

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(DIV(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

POL(p(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳FwdInst

           →DP Problem 7

             ↳Nar

...

               →DP Problem 10

                 ↳Polynomial Ordering

Dependency Pair:

DIV(s(s(s(X'))), s(s(s(Y')))) -> DIV(p(p(minus(X', Y'))), s(s(s(Y'))))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

DIV(s(s(s(X'))), s(s(s(Y')))) -> DIV(p(p(minus(X', Y'))), s(s(s(Y'))))

Additionally, the following usable rules for innermost can be oriented:

p(s(X)) -> X
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 1

POL(DIV(x₁, x₂)) = 1 + x₁

POL(minus(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(p(x₁)) = x₁

resulting in one new DP problem.

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

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

POL(0)	= 0
POL(DIV(x₁, x₂))	= 1 + x₁
POL(s(x₁))	= 1 + x₁