Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACK(s(x), 0) -> ACK(x, s(0))
ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
ACK(s(x), s(y)) -> ACK(s(x), y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

ACK(s(x), s(y)) -> ACK(s(x), y)
ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
ACK(s(x), 0) -> ACK(x, s(0))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))

two new Dependency Pairs are created:

ACK(s(x''), s(0)) -> ACK(x'', ack(x'', s(0)))
ACK(s(x''), s(s(y''))) -> ACK(x'', ack(x'', ack(s(x''), y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

ACK(s(x''), s(s(y''))) -> ACK(x'', ack(x'', ack(s(x''), y'')))
ACK(s(x''), s(0)) -> ACK(x'', ack(x'', s(0)))
ACK(s(x), 0) -> ACK(x, s(0))
ACK(s(x), s(y)) -> ACK(s(x), y)

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

ACK(s(x''), s(0)) -> ACK(x'', ack(x'', s(0)))

two new Dependency Pairs are created:

ACK(s(0), s(0)) -> ACK(0, s(s(0)))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(s(x'), 0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Rewriting Transformation

Dependency Pairs:

ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(s(x'), 0)))
ACK(s(x), s(y)) -> ACK(s(x), y)
ACK(s(x), 0) -> ACK(x, s(0))
ACK(s(x''), s(s(y''))) -> ACK(x'', ack(x'', ack(s(x''), y'')))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(s(x'), 0)))

one new Dependency Pair is created:

ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

ACK(s(x''), s(s(y''))) -> ACK(x'', ack(x'', ack(s(x''), y'')))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))
ACK(s(x), 0) -> ACK(x, s(0))
ACK(s(x), s(y)) -> ACK(s(x), y)

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

ACK(s(x''), s(s(y''))) -> ACK(x'', ack(x'', ack(s(x''), y'')))

three new Dependency Pairs are created:

ACK(s(0), s(s(y'''))) -> ACK(0, s(ack(s(0), y''')))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(x), s(y)) -> ACK(s(x), y)
ACK(s(x), 0) -> ACK(x, s(0))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

ACK(s(x), 0) -> ACK(x, s(0))

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pairs:

ACK(s(s(s(x'''))), 0) -> ACK(s(s(x''')), s(0))
ACK(s(s(x'')), 0) -> ACK(s(x''), s(0))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))
ACK(s(x), s(y)) -> ACK(s(x), y)
ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

ACK(s(x), s(y)) -> ACK(s(x), y)

six new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

Dependency Pairs:

ACK(s(x'), s(s(s(s(y'''))))) -> ACK(s(x'), s(s(s(y'''))))
ACK(s(x'), s(s(s(0)))) -> ACK(s(x'), s(s(0)))
ACK(s(s(x''')), s(s(0))) -> ACK(s(s(x''')), s(0))
ACK(s(x''), s(s(y''))) -> ACK(s(x''), s(y''))
ACK(s(s(s(x'''''))), s(0)) -> ACK(s(s(s(x'''''))), 0)
ACK(s(s(x'''')), s(0)) -> ACK(s(s(x'''')), 0)
ACK(s(s(x'')), 0) -> ACK(s(x''), s(0))
ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))
ACK(s(s(s(x'''))), 0) -> ACK(s(s(x''')), s(0))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

Dependency Pairs:

ACK(s(x'), s(s(s(0)))) -> ACK(s(x'), s(s(0)))
ACK(s(s(x''')), s(s(0))) -> ACK(s(s(x''')), s(0))
ACK(s(x''), s(s(y''))) -> ACK(s(x''), s(y''))
ACK(s(s(s(s(x''''''')))), 0) -> ACK(s(s(s(x'''''''))), s(0))
ACK(s(s(s(x''''''))), 0) -> ACK(s(s(x'''''')), s(0))
ACK(s(s(s(x'''''))), s(0)) -> ACK(s(s(s(x'''''))), 0)
ACK(s(s(s(x''''))), 0) -> ACK(s(s(x'''')), s(0))
ACK(s(s(x'''')), s(0)) -> ACK(s(s(x'''')), 0)
ACK(s(s(s(x'''))), 0) -> ACK(s(s(x''')), s(0))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))
ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))
ACK(s(x'), s(s(s(s(y'''))))) -> ACK(s(x'), s(s(s(y'''))))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

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

nine new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

Dependency Pairs:

ACK(s(x''''), s(s(s(s(s(y''''')))))) -> ACK(s(x''''), s(s(s(s(y''''')))))
ACK(s(x''''), s(s(s(s(0))))) -> ACK(s(x''''), s(s(s(0))))
ACK(s(s(x''''')), s(s(s(0)))) -> ACK(s(s(x''''')), s(s(0)))
ACK(s(x''''), s(s(s(y'''')))) -> ACK(s(x''''), s(s(y'''')))
ACK(s(x'''), s(s(s(s(y''''))))) -> ACK(s(x'''), s(s(s(y''''))))
ACK(s(s(s(x'''''''))), s(s(0))) -> ACK(s(s(s(x'''''''))), s(0))
ACK(s(s(x'''''')), s(s(0))) -> ACK(s(s(x'''''')), s(0))
ACK(s(s(x'''')), s(s(0))) -> ACK(s(s(x'''')), s(0))
ACK(s(x'''), s(s(s(0)))) -> ACK(s(x'''), s(s(0)))
ACK(s(x'), s(s(s(s(y'''))))) -> ACK(s(x'), s(s(s(y'''))))
ACK(s(s(x''')), s(s(0))) -> ACK(s(s(x''')), s(0))
ACK(s(s(s(s(x''''''')))), 0) -> ACK(s(s(s(x'''''''))), s(0))
ACK(s(s(s(x''''''))), 0) -> ACK(s(s(x'''''')), s(0))
ACK(s(s(s(x'''''))), s(0)) -> ACK(s(s(s(x'''''))), 0)
ACK(s(s(s(x''''))), 0) -> ACK(s(s(x'''')), s(0))
ACK(s(s(x'''')), s(0)) -> ACK(s(s(x'''')), 0)
ACK(s(s(s(x'''))), 0) -> ACK(s(s(x''')), s(0))
ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(x'), s(s(s(0)))) -> ACK(s(x'), s(s(0)))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 10

                 ↳Argument Filtering and Ordering

Dependency Pairs:

ACK(s(x''''), s(s(s(s(0))))) -> ACK(s(x''''), s(s(s(0))))
ACK(s(s(x''''')), s(s(s(0)))) -> ACK(s(s(x''''')), s(s(0)))
ACK(s(x''''), s(s(s(y'''')))) -> ACK(s(x''''), s(s(y'''')))
ACK(s(x'''), s(s(s(s(y''''))))) -> ACK(s(x'''), s(s(s(y''''))))
ACK(s(x'''), s(s(s(0)))) -> ACK(s(x'''), s(s(0)))
ACK(s(x'), s(s(s(s(y'''))))) -> ACK(s(x'), s(s(s(y'''))))
ACK(s(s(s(x'''''''))), s(s(0))) -> ACK(s(s(s(x'''''''))), s(0))
ACK(s(s(x'''''')), s(s(0))) -> ACK(s(s(x'''''')), s(0))
ACK(s(s(x'''')), s(s(0))) -> ACK(s(s(x'''')), s(0))
ACK(s(x'), s(s(s(0)))) -> ACK(s(x'), s(s(0)))
ACK(s(s(x''')), s(s(0))) -> ACK(s(s(x''')), s(0))
ACK(s(s(s(s(x''''''''')))), s(0)) -> ACK(s(s(s(s(x''''''''')))), 0)
ACK(s(s(s(s(x''''''')))), 0) -> ACK(s(s(s(x'''''''))), s(0))
ACK(s(s(s(x''''''''))), s(0)) -> ACK(s(s(s(x''''''''))), 0)
ACK(s(s(s(x''''''))), 0) -> ACK(s(s(x'''''')), s(0))
ACK(s(s(s(x''''''))), s(0)) -> ACK(s(s(s(x''''''))), 0)
ACK(s(s(s(x''''))), 0) -> ACK(s(s(x'''')), s(0))
ACK(s(s(s(x'''''))), s(0)) -> ACK(s(s(s(x'''''))), 0)
ACK(s(s(s(x'''))), 0) -> ACK(s(s(x''')), s(0))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))
ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))
ACK(s(x''''), s(s(s(s(s(y''''')))))) -> ACK(s(x''''), s(s(s(s(y''''')))))

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

ACK(s(x''''), s(s(s(s(0))))) -> ACK(s(x''''), s(s(s(0))))
ACK(s(s(x''''')), s(s(s(0)))) -> ACK(s(s(x''''')), s(s(0)))
ACK(s(x''''), s(s(s(y'''')))) -> ACK(s(x''''), s(s(y'''')))
ACK(s(x'''), s(s(s(s(y''''))))) -> ACK(s(x'''), s(s(s(y''''))))
ACK(s(x'''), s(s(s(0)))) -> ACK(s(x'''), s(s(0)))
ACK(s(x'), s(s(s(s(y'''))))) -> ACK(s(x'), s(s(s(y'''))))
ACK(s(s(s(x'''''''))), s(s(0))) -> ACK(s(s(s(x'''''''))), s(0))
ACK(s(s(x'''''')), s(s(0))) -> ACK(s(s(x'''''')), s(0))
ACK(s(s(x'''')), s(s(0))) -> ACK(s(s(x'''')), s(0))
ACK(s(x'), s(s(s(0)))) -> ACK(s(x'), s(s(0)))
ACK(s(s(x''')), s(s(0))) -> ACK(s(s(x''')), s(0))
ACK(s(s(s(s(x''''''''')))), s(0)) -> ACK(s(s(s(s(x''''''''')))), 0)
ACK(s(s(s(s(x''''''')))), 0) -> ACK(s(s(s(x'''''''))), s(0))
ACK(s(s(s(x''''''''))), s(0)) -> ACK(s(s(s(x''''''''))), 0)
ACK(s(s(s(x''''''))), 0) -> ACK(s(s(x'''''')), s(0))
ACK(s(s(s(x''''''))), s(0)) -> ACK(s(s(s(x''''''))), 0)
ACK(s(s(s(x''''))), 0) -> ACK(s(s(x'''')), s(0))
ACK(s(s(s(x'''''))), s(0)) -> ACK(s(s(s(x'''''))), 0)
ACK(s(s(s(x'''))), 0) -> ACK(s(s(x''')), s(0))
ACK(s(x'''), s(s(0))) -> ACK(x''', ack(x''', ack(x''', s(0))))
ACK(s(s(x')), s(0)) -> ACK(s(x'), ack(x', ack(x', s(0))))
ACK(s(x'''), s(s(s(y')))) -> ACK(x''', ack(x''', ack(x''', ack(s(x'''), y'))))
ACK(s(x''''), s(s(s(s(s(y''''')))))) -> ACK(s(x''''), s(s(s(s(y''''')))))

The following usable rules for innermost can be oriented:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

ACK > ack > s

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

ACK(x₁, x₂) -> ACK(x₁, x₂)
s(x₁) -> s(x₁)
ack(x₁, x₂) -> ack(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 11

                 ↳Dependency Graph

Dependency Pair:

Rules:

ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, ack(s(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes