Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

G(x, s(y)) -> G(f(x, y), 0)
G(s(x), y) -> G(f(x, y), 0)
G(f(x, y), 0) -> G(x, 0)
G(f(x, y), 0) -> G(y, 0)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Instantiation Transformation

Dependency Pairs:

G(f(x, y), 0) -> G(y, 0)
G(s(x), y) -> G(f(x, y), 0)
G(f(x, y), 0) -> G(x, 0)

Rules:

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

Strategy:

innermost

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

G(s(x), y) -> G(f(x, y), 0)

two new Dependency Pairs are created:

G(s(x''), 0) -> G(f(x'', 0), 0)
G(s(x'), 0) -> G(f(x', 0), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

G(s(x'), 0) -> G(f(x', 0), 0)
G(s(x''), 0) -> G(f(x'', 0), 0)
G(f(x, y), 0) -> G(x, 0)
G(f(x, y), 0) -> G(y, 0)

Rules:

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

Strategy:

innermost

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

G(f(x, y), 0) -> G(x, 0)

three new Dependency Pairs are created:

G(f(f(x'', y''), y), 0) -> G(f(x'', y''), 0)
G(f(s(x''''), y), 0) -> G(s(x''''), 0)
G(f(s(x'''), y), 0) -> G(s(x'''), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(f(s(x'''), y), 0) -> G(s(x'''), 0)
G(f(s(x''''), y), 0) -> G(s(x''''), 0)
G(f(f(x'', y''), y), 0) -> G(f(x'', y''), 0)
G(s(x''), 0) -> G(f(x'', 0), 0)
G(f(x, y), 0) -> G(y, 0)
G(s(x'), 0) -> G(f(x', 0), 0)

Rules:

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

Strategy:

innermost

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

G(f(x, y), 0) -> G(y, 0)

six new Dependency Pairs are created:

G(f(x, f(x'', y'')), 0) -> G(f(x'', y''), 0)
G(f(x, s(x'''')), 0) -> G(s(x''''), 0)
G(f(x, s(x''')), 0) -> G(s(x'''), 0)
G(f(x, f(f(x'''', y''''), y'')), 0) -> G(f(f(x'''', y''''), y''), 0)
G(f(x, f(s(x''''''), y'')), 0) -> G(f(s(x''''''), y''), 0)
G(f(x, f(s(x'''''), y'')), 0) -> G(f(s(x'''''), y''), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(f(x, f(s(x'''''), y'')), 0) -> G(f(s(x'''''), y''), 0)
G(f(x, f(s(x''''''), y'')), 0) -> G(f(s(x''''''), y''), 0)
G(f(x, f(f(x'''', y''''), y'')), 0) -> G(f(f(x'''', y''''), y''), 0)
G(f(x, s(x''')), 0) -> G(s(x'''), 0)
G(f(x, s(x'''')), 0) -> G(s(x''''), 0)
G(f(x, f(x'', y'')), 0) -> G(f(x'', y''), 0)
G(s(x'), 0) -> G(f(x', 0), 0)
G(f(s(x''''), y), 0) -> G(s(x''''), 0)
G(f(f(x'', y''), y), 0) -> G(f(x'', y''), 0)
G(s(x''), 0) -> G(f(x'', 0), 0)
G(f(s(x'''), y), 0) -> G(s(x'''), 0)

Rules:

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

Strategy:

innermost

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

G(s(x''), 0) -> G(f(x'', 0), 0)

three new Dependency Pairs are created:

G(s(f(x'''', y'''')), 0) -> G(f(f(x'''', y''''), 0), 0)
G(s(s(x'''''')), 0) -> G(f(s(x''''''), 0), 0)
G(s(s(x''''')), 0) -> G(f(s(x'''''), 0), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(f(x, f(s(x''''''), y'')), 0) -> G(f(s(x''''''), y''), 0)
G(f(x, f(f(x'''', y''''), y'')), 0) -> G(f(f(x'''', y''''), y''), 0)
G(f(x, s(x''')), 0) -> G(s(x'''), 0)
G(f(x, s(x'''')), 0) -> G(s(x''''), 0)
G(f(x, f(x'', y'')), 0) -> G(f(x'', y''), 0)
G(s(s(x''''')), 0) -> G(f(s(x'''''), 0), 0)
G(s(s(x'''''')), 0) -> G(f(s(x''''''), 0), 0)
G(s(f(x'''', y'''')), 0) -> G(f(f(x'''', y''''), 0), 0)
G(f(s(x'''), y), 0) -> G(s(x'''), 0)
G(f(f(x'', y''), y), 0) -> G(f(x'', y''), 0)
G(s(x'), 0) -> G(f(x', 0), 0)
G(f(s(x''''), y), 0) -> G(s(x''''), 0)
G(f(x, f(s(x'''''), y'')), 0) -> G(f(s(x'''''), y''), 0)

Rules:

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

Strategy:

innermost

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

G(s(x'), 0) -> G(f(x', 0), 0)

three new Dependency Pairs are created:

G(s(f(x'''', y'''')), 0) -> G(f(f(x'''', y''''), 0), 0)
G(s(s(x'''''')), 0) -> G(f(s(x''''''), 0), 0)
G(s(s(x''''')), 0) -> G(f(s(x'''''), 0), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

G(f(x, f(s(x'''''), y'')), 0) -> G(f(s(x'''''), y''), 0)
G(f(x, f(f(x'''', y''''), y'')), 0) -> G(f(f(x'''', y''''), y''), 0)
G(f(x, s(x''')), 0) -> G(s(x'''), 0)
G(f(x, s(x'''')), 0) -> G(s(x''''), 0)
G(f(x, f(x'', y'')), 0) -> G(f(x'', y''), 0)
G(s(s(x''''')), 0) -> G(f(s(x'''''), 0), 0)
G(s(s(x'''''')), 0) -> G(f(s(x''''''), 0), 0)
G(s(f(x'''', y'''')), 0) -> G(f(f(x'''', y''''), 0), 0)
G(s(s(x''''')), 0) -> G(f(s(x'''''), 0), 0)
G(s(s(x'''''')), 0) -> G(f(s(x''''''), 0), 0)
G(f(s(x'''), y), 0) -> G(s(x'''), 0)
G(f(f(x'', y''), y), 0) -> G(f(x'', y''), 0)
G(s(f(x'''', y'''')), 0) -> G(f(f(x'''', y''''), 0), 0)
G(f(s(x''''), y), 0) -> G(s(x''''), 0)
G(f(x, f(s(x''''''), y'')), 0) -> G(f(s(x''''''), y''), 0)

Rules:

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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:02 minutes