Termination Proof using AProVE

Term Rewriting System R:
[x, y]
f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Rw

Dependency Pair:

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

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

         ↳Rw

Dependency Pair:

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

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Rw

Dependency Pair:

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

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

+'(x₁, x₂) -> +'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Rw

Dependency Pair:

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rewriting Transformation

Dependency Pairs:

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

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

G(s(0), y') -> G(0, s(y'))
G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))
G(s(x), y) -> G(x, s(+(y, x)))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

G(s(0), y') -> G(0, s(y'))
G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))
G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))

two new Dependency Pairs are created:

G(s(s(0)), y0') -> G(s(0), s(s(y0')))
G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))
G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(y'')), y0) -> G(s(y''), s(s(+(y0, y''))))

two new Dependency Pairs are created:

G(s(s(0)), y0') -> G(s(0), s(s(y0')))
G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))
G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))

two new Dependency Pairs are created:

G(s(s(s(0))), y0'') -> G(s(s(0)), s(s(s(y0''))))
G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))
G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(s(y'))), y0') -> G(s(s(y')), s(s(s(+(y0', y')))))

two new Dependency Pairs are created:

G(s(s(s(0))), y0'') -> G(s(s(0)), s(s(s(y0''))))
G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))
G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))

two new Dependency Pairs are created:

G(s(s(s(s(0)))), y0''') -> G(s(s(s(0))), s(s(s(s(y0''')))))
G(s(s(s(s(s(y'))))), y0''') -> G(s(s(s(s(y')))), s(s(s(s(s(+(y0''', y')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 13

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(s(s(s(y'))))), y0''') -> G(s(s(s(s(y')))), s(s(s(s(s(+(y0''', y')))))))
G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(s(s(y'')))), y0'') -> G(s(s(s(y''))), s(s(s(s(+(y0'', y''))))))

two new Dependency Pairs are created:

G(s(s(s(s(0)))), y0''') -> G(s(s(s(0))), s(s(s(s(y0''')))))
G(s(s(s(s(s(y'))))), y0''') -> G(s(s(s(s(y')))), s(s(s(s(s(+(y0''', y')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 14

                 ↳Narrowing Transformation

Dependency Pairs:

G(s(s(s(s(s(y'))))), y0''') -> G(s(s(s(s(y')))), s(s(s(s(s(+(y0''', y')))))))
G(s(s(s(s(s(y'))))), y0''') -> G(s(s(s(s(y')))), s(s(s(s(s(+(y0''', y')))))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

G(s(s(s(s(s(y'))))), y0''') -> G(s(s(s(s(y')))), s(s(s(s(s(+(y0''', y')))))))

two new Dependency Pairs are created:

G(s(s(s(s(s(0))))), y0'''') -> G(s(s(s(s(0)))), s(s(s(s(s(y0''''))))))
G(s(s(s(s(s(s(y'')))))), y0'''') -> G(s(s(s(s(s(y''))))), s(s(s(s(s(s(+(y0'''', y''))))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 15

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

G(s(s(s(s(s(s(y'')))))), y0'''') -> G(s(s(s(s(s(y''))))), s(s(s(s(s(s(+(y0'''', y''))))))))
G(s(s(s(s(s(y'))))), y0''') -> G(s(s(s(s(y')))), s(s(s(s(s(+(y0''', y')))))))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes