Termination Proof using AProVE

Term Rewriting System R:
[m, n, r]
p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

P(m, n, s(r)) -> P(m, r, n)
P(m, s(n), 0) -> P(0, n, m)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

Dependency Pairs:

P(m, s(n), 0) -> P(0, n, m)
P(m, n, s(r)) -> P(m, r, n)

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

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

P(m, n, s(r)) -> P(m, r, n)

two new Dependency Pairs are created:

P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))
P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)
P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))
P(m, s(n), 0) -> P(0, n, m)

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

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

P(m, s(n), 0) -> P(0, n, m)

three new Dependency Pairs are created:

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(s(s(n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pair:

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

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

P(0, s(s(n'')), 0) -> P(0, s(n''), 0)

one new Dependency Pair is created:

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

Dependency Pair:

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

The following dependency pair can be strictly oriented:

P(0, s(s(s(n''''))), 0) -> P(0, s(s(n'''')), 0)

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(P(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

P(x₁, x₂, x₃) -> P(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

P(s(s(n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))
P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

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

P(m'', s(r''), s(r0)) -> P(m'', r0, s(r''))

two new Dependency Pairs are created:

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))
P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)
P(s(s(n'''')), s(0), 0) -> P(0, 0, s(s(n'''')))

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

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

P(m'', 0, s(s(n''))) -> P(m'', s(n''), 0)

two new Dependency Pairs are created:

P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(s(s(n'''''')), 0, s(s(0))) -> P(s(s(n'''''')), s(0), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

Dependency Pairs:

P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))
P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))
P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

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

P(s(r0''), s(s(r'''')), 0) -> P(0, s(r''''), s(r0''))

two new Dependency Pairs are created:

P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))
P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

Dependency Pairs:

P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))
P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))
P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))
P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))

Rules:

p(m, n, s(r)) -> p(m, r, n)
p(m, s(n), 0) -> p(0, n, m)
p(m, 0, 0) -> m

Strategy:

innermost

The following dependency pairs can be strictly oriented:

P(s(0), s(s(s(n''''''))), 0) -> P(0, s(s(n'''''')), s(0))
P(s(s(r'''''''')), s(s(r''''')), 0) -> P(0, s(r'''''), s(s(r'''''''')))
P(s(r0''''), 0, s(s(s(r'''''')))) -> P(s(r0''''), s(s(r'''''')), 0)
P(m'''', s(s(n'''')), s(0)) -> P(m'''', 0, s(s(n'''')))
P(m'''', s(r''''), s(s(r'''''))) -> P(m'''', s(r'''''), s(r''''))

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(P(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

P(x₁, x₂, x₃) -> P(x₁, x₂, x₃)
s(x₁) -> s(x₁)

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

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