Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(x, s(y)) -> +'(x, y)
+'(s(x), y) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
F(g(f(x))) -> F(h(s(0), x))
F(g(h(x, y))) -> F(h(s(x), y))
F(h(x, h(y, z))) -> F(h(+(x, y), z))
F(h(x, h(y, z))) -> +'(x, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳AFS

...

               →DP Problem 4

                 ↳Dependency Graph

       →DP Problem 2

         ↳AFS

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(h(x, h(y, z))) -> F(h(+(x, y), z))

F(x₁) -> F(x₁)
h(x₁, x₂) -> h(x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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