Termination Proof using AProVE

Term Rewriting System R:
[y, z, x]
quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

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

QUOT(s(x), s(y), z) -> QUOT(x, y, z)

two new Dependency Pairs are created:

QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(s(x''), s(0), s(z'')) -> QUOT(x'', 0, s(z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(s(x''), s(0), s(z'')) -> QUOT(x'', 0, s(z''))
QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

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

QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))

two new Dependency Pairs are created:

QUOT(s(s(x'''')), 0, s(s(y''''))) -> QUOT(s(s(x'''')), s(s(y'''')), s(s(y'''')))
QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))
QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(s(s(x'''')), 0, s(s(y''''))) -> QUOT(s(s(x'''')), s(s(y'''')), s(s(y'''')))
QUOT(s(x''), s(0), s(z'')) -> QUOT(x'', 0, s(z''))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(s(s(x'''')), s(s(0)), s(z'''')) -> QUOT(s(x''''), s(0), s(z''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(s(s(x'''')), s(s(0)), s(z'''')) -> QUOT(s(x''''), s(0), s(z''''))
QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(s(s(x'''')), 0, s(s(y''''))) -> QUOT(s(s(x'''')), s(s(y'''')), s(s(y'''')))
QUOT(s(x''), s(0), s(z'')) -> QUOT(x'', 0, s(z''))
QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

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

QUOT(s(x''), s(0), s(z'')) -> QUOT(x'', 0, s(z''))

two new Dependency Pairs are created:

QUOT(s(s(s(x''''''))), s(0), s(s(y''''''))) -> QUOT(s(s(x'''''')), 0, s(s(y'''''')))
QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))
QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

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

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

one new Dependency Pair is created:

QUOT(s(s(x'''''''')), 0, s(0)) -> QUOT(s(s(x'''''''')), s(0), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

QUOT(s(s(x'''''''')), 0, s(0)) -> QUOT(s(s(x'''''''')), s(0), s(0))
QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂, x₃)) = x₁ + x₃

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

QUOT(s(s(x'''''''')), 0, s(0)) -> QUOT(s(s(x'''''''')), s(0), s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(s(s(x'''')), 0, s(s(y''''))) -> QUOT(s(s(x'''')), s(s(y'''')), s(s(y'''')))
QUOT(s(s(s(x''''''))), s(0), s(s(y''''''))) -> QUOT(s(s(x'''''')), 0, s(s(y'''''')))
QUOT(s(s(x'''')), s(s(0)), s(z'''')) -> QUOT(s(x''''), s(0), s(z''''))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

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

QUOT(s(s(x'''')), 0, s(s(y''''))) -> QUOT(s(s(x'''')), s(s(y'''')), s(s(y'''')))

two new Dependency Pairs are created:

QUOT(s(s(s(x''''''))), 0, s(s(s(y'''''')))) -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), s(s(s(y''''''))))
QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

Dependency Pairs:

QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))
QUOT(s(s(s(x''''''))), 0, s(s(s(y'''''')))) -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), s(s(s(y''''''))))
QUOT(s(s(s(x''''''))), s(0), s(s(y''''''))) -> QUOT(s(s(x'''''')), 0, s(s(y'''''')))
QUOT(s(s(x'''')), s(s(0)), s(z'''')) -> QUOT(s(x''''), s(0), s(z''''))
QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

QUOT(s(s(s(x''''''))), s(0), s(s(y''''''))) -> QUOT(s(s(x'''''')), 0, s(s(y'''''')))
QUOT(s(s(x'''')), s(s(0)), s(z'''')) -> QUOT(s(x''''), s(0), s(z''''))
QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂, x₃)) = 1 + x₁

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

Dependency Pairs:

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(QUOT(x₁, x₂, x₃))	= x₁ + x₃
POL(0)	= 0
POL(s(x₁))	= 1 + x₁