Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
F(s(x)) -> MINUS(s(x), g(f(x)))
F(s(x)) -> G(f(x))
F(s(x)) -> F(x)
G(s(x)) -> MINUS(s(x), f(g(x)))
G(s(x)) -> F(g(x))
G(s(x)) -> G(x)

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)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

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)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

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)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

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 w.r.t. to the implicit AFS 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)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
F(s(x)) -> G(f(x))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(x)) -> G(f(x))

two new Dependency Pairs are created:

F(s(0)) -> G(s(0))
F(s(s(x''))) -> G(minus(s(x''), g(f(x''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

F(s(s(x''))) -> G(minus(s(x''), g(f(x''))))
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
G(s(x)) -> G(x)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

G(s(x)) -> F(g(x))

two new Dependency Pairs are created:

G(s(0)) -> F(0)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(s(x''))) -> G(minus(s(x''), g(f(x''))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(x''))) -> G(minus(s(x''), g(f(x''))))

two new Dependency Pairs are created:

F(s(s(0))) -> G(minus(s(0), g(s(0))))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(s(0))) -> G(minus(s(0), g(s(0))))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(0)) -> G(s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(0))) -> G(minus(s(0), g(s(0))))

one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(s(0), f(g(0)))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(0))) -> G(minus(s(0), minus(s(0), f(g(0)))))
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(0))) -> G(minus(s(0), minus(s(0), f(g(0)))))

one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(s(0), f(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(0))) -> G(minus(s(0), minus(s(0), f(0))))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(0)) -> G(s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(0))) -> G(minus(s(0), minus(s(0), f(0))))

one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(s(0), s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(0))) -> G(minus(s(0), minus(s(0), s(0))))
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(0))) -> G(minus(s(0), minus(s(0), s(0))))

one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), minus(0, 0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(0))) -> G(minus(s(0), minus(0, 0)))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(0)) -> G(s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(0))) -> G(minus(s(0), minus(0, 0)))

one new Dependency Pair is created:

F(s(s(0))) -> G(minus(s(0), 0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 13

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(0))) -> G(minus(s(0), 0))
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(0))) -> G(minus(s(0), 0))

one new Dependency Pair is created:

F(s(s(0))) -> G(s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 14

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(s(0))) -> G(s(0))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(x)) -> F(x)
G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))
G(s(x)) -> G(x)
F(s(0)) -> G(s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

G(s(s(x''))) -> F(minus(s(x''), f(g(x''))))

two new Dependency Pairs are created:

G(s(s(0))) -> F(minus(s(0), f(0)))
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 15

                 ↳Rewriting Transformation

Dependency Pairs:

G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(0))) -> F(minus(s(0), f(0)))
G(s(x)) -> G(x)
F(s(s(0))) -> G(s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

G(s(s(0))) -> F(minus(s(0), f(0)))

one new Dependency Pair is created:

G(s(s(0))) -> F(minus(s(0), s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 16

                 ↳Rewriting Transformation

Dependency Pairs:

F(s(s(0))) -> G(s(0))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
G(s(s(0))) -> F(minus(s(0), s(0)))
G(s(x)) -> G(x)
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

G(s(s(0))) -> F(minus(s(0), s(0)))

one new Dependency Pair is created:

G(s(s(0))) -> F(minus(0, 0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 17

                 ↳Rewriting Transformation

Dependency Pairs:

G(s(s(0))) -> F(minus(0, 0))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
G(s(x)) -> G(x)
F(s(s(0))) -> G(s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

G(s(s(0))) -> F(minus(0, 0))

one new Dependency Pair is created:

G(s(s(0))) -> F(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(0))) -> G(s(0))
F(s(0)) -> G(s(0))
F(s(x)) -> F(x)
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
G(s(x)) -> G(x)
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(x)) -> F(x)

four new Dependency Pairs are created:

F(s(s(x''))) -> F(s(x''))
F(s(s(0))) -> F(s(0))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
F(s(s(s(0)))) -> F(s(s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 19

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(s(0)))) -> F(s(s(0)))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
F(s(s(0))) -> F(s(0))
F(s(s(x''))) -> F(s(x''))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(0)) -> G(s(0))
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
G(s(x)) -> G(x)
F(s(s(0))) -> G(s(0))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

G(s(x)) -> G(x)

two new Dependency Pairs are created:

G(s(s(x''))) -> G(s(x''))
G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 20

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))
G(s(s(x''))) -> G(s(x''))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(s(x''))) -> F(s(x''))
F(s(s(s(0)))) -> F(s(s(0)))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

F(s(s(x''))) -> F(s(x''))

four new Dependency Pairs are created:

F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(x'''')))) -> F(s(s(x'''')))
F(s(s(s(s(s(x''''')))))) -> F(s(s(s(s(x''''')))))
F(s(s(s(s(0))))) -> F(s(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 21

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(s(s(0))))) -> F(s(s(s(0))))
F(s(s(s(s(s(x''''')))))) -> F(s(s(s(s(x''''')))))
F(s(s(s(x'''')))) -> F(s(s(x'''')))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
G(s(s(x''))) -> G(s(x''))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

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

G(s(s(x''))) -> G(s(x''))

three new Dependency Pairs are created:

G(s(s(s(s(x''''))))) -> G(s(s(s(x''''))))
G(s(s(s(x'''')))) -> G(s(s(x'''')))
G(s(s(s(s(s(x''''')))))) -> G(s(s(s(s(x''''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 22

                 ↳Polynomial Ordering

Dependency Pairs:

G(s(s(s(s(s(x''''')))))) -> G(s(s(s(s(x''''')))))
G(s(s(s(x'''')))) -> G(s(s(x'''')))
G(s(s(s(s(x''''))))) -> G(s(s(s(x''''))))
G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))
F(s(s(s(s(s(x''''')))))) -> F(s(s(s(s(x''''')))))
F(s(s(s(x'''')))) -> F(s(s(x'''')))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(s(s(s(0))))) -> F(s(s(s(0))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

G(s(s(s(s(s(x''''')))))) -> G(s(s(s(s(x''''')))))
G(s(s(s(x'''')))) -> G(s(s(x'''')))
G(s(s(s(s(x''''))))) -> G(s(s(s(x''''))))
G(s(s(s(s(x'''))))) -> G(s(s(s(x'''))))
F(s(s(s(s(s(x''''')))))) -> F(s(s(s(s(x''''')))))
F(s(s(s(x'''')))) -> F(s(s(x'''')))
F(s(s(s(s(x''''))))) -> F(s(s(s(x''''))))
F(s(s(s(s(x'''))))) -> F(s(s(s(x'''))))
G(s(s(s(x')))) -> F(minus(s(s(x')), f(minus(s(x'), f(g(x'))))))
F(s(s(s(x')))) -> G(minus(s(s(x')), g(minus(s(x'), g(f(x'))))))
F(s(s(s(s(0))))) -> F(s(s(s(0))))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(g(x₁)) = 0

POL(G(x₁)) = x₁

POL(minus(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(f(x₁)) = 0

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 23

                 ↳Dependency Graph

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

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

POL(0)	= 0
POL(g(x₁))	= 0
POL(G(x₁))	= x₁
POL(minus(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁
POL(f(x₁))	= 0
POL(F(x₁))	= x₁