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

                 ↳Polynomial 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 w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳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

                 ↳Polynomial Ordering

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

The following dependency pairs can be strictly oriented:

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''))))

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(0) = 0

POL(G(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

POL(+(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Dependency Graph

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.

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

POL(0)	= 0
POL(G(x₁, x₂))	= 1 + x₁
POL(s(x₁))	= 1 + x₁
POL(+(x₁, x₂))	= 0