Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(g(x), a) -> F(x, g(a))
F(g(x), g(y)) -> H(g(y), x, g(y))
H(g(x), y, z) -> F(y, h(x, y, z))
H(g(x), y, z) -> H(x, y, z)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

Dependency Pairs:

H(g(x), y, z) -> H(x, y, z)
H(g(x), y, z) -> F(y, h(x, y, z))
F(g(x), g(y)) -> H(g(y), x, g(y))
F(g(x), a) -> F(x, g(a))

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(g(x), a) -> F(x, g(a))

Additionally, the following usable rules for innermost can be oriented:

h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z
f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(g(x₁)) = 0

POL(h(x₁, x₂, x₃)) = x₃

POL(a) = 1

POL(H(x₁, x₂, x₃)) = x₃

POL(f(x₁, x₂)) = 0

POL(F(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

H(g(x), y, z) -> H(x, y, z)
H(g(x), y, z) -> F(y, h(x, y, z))
F(g(x), g(y)) -> H(g(y), x, g(y))

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

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

H(g(x), y, z) -> F(y, h(x, y, z))

two new Dependency Pairs are created:

H(g(g(x'')), y'', z'') -> F(y'', f(y'', h(x'', y'', z'')))
H(g(a), y'', z'') -> F(y'', z'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

H(g(a), y'', z'') -> F(y'', z'')
F(g(x), g(y)) -> H(g(y), x, g(y))
H(g(g(x'')), y'', z'') -> F(y'', f(y'', h(x'', y'', z'')))
H(g(x), y, z) -> H(x, y, z)

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

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

H(g(x), y, z) -> H(x, y, z)

three new Dependency Pairs are created:

H(g(g(x'')), y'', z'') -> H(g(x''), y'', z'')
H(g(g(g(x''''))), y', z') -> H(g(g(x'''')), y', z')
H(g(g(a)), y', z') -> H(g(a), y', z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

H(g(g(a)), y', z') -> H(g(a), y', z')
H(g(g(g(x''''))), y', z') -> H(g(g(x'''')), y', z')
H(g(g(x'')), y'', z'') -> H(g(x''), y'', z'')
H(g(g(x'')), y'', z'') -> F(y'', f(y'', h(x'', y'', z'')))
F(g(x), g(y)) -> H(g(y), x, g(y))
H(g(a), y'', z'') -> F(y'', z'')

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

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

F(g(x), g(y)) -> H(g(y), x, g(y))

four new Dependency Pairs are created:

F(g(x'), g(g(x''''))) -> H(g(g(x'''')), x', g(g(x'''')))
F(g(x'), g(a)) -> H(g(a), x', g(a))
F(g(x'), g(g(g(x'''''')))) -> H(g(g(g(x''''''))), x', g(g(g(x''''''))))
F(g(x'), g(g(a))) -> H(g(g(a)), x', g(g(a)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(g(x'), g(g(a))) -> H(g(g(a)), x', g(g(a)))
H(g(g(g(x''''))), y', z') -> H(g(g(x'''')), y', z')
H(g(g(x'')), y'', z'') -> H(g(x''), y'', z'')
F(g(x'), g(g(g(x'''''')))) -> H(g(g(g(x''''''))), x', g(g(g(x''''''))))
F(g(x'), g(a)) -> H(g(a), x', g(a))
H(g(g(x'')), y'', z'') -> F(y'', f(y'', h(x'', y'', z'')))
F(g(x'), g(g(x''''))) -> H(g(g(x'''')), x', g(g(x'''')))
H(g(a), y'', z'') -> F(y'', z'')
H(g(g(a)), y', z') -> H(g(a), y', z')

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

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

H(g(a), y'', z'') -> F(y'', z'')

four new Dependency Pairs are created:

H(g(a), g(x'''), g(g(x''''''))) -> F(g(x'''), g(g(x'''''')))
H(g(a), g(x'''), g(a)) -> F(g(x'''), g(a))
H(g(a), g(x'''), g(g(g(x'''''''')))) -> F(g(x'''), g(g(g(x''''''''))))
H(g(a), g(x'''), g(g(a))) -> F(g(x'''), g(g(a)))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Polynomial Ordering

Dependency Pairs:

F(g(x'), g(a)) -> H(g(a), x', g(a))
H(g(a), g(x'''), g(a)) -> F(g(x'''), g(a))

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(g(x'), g(a)) -> H(g(a), x', g(a))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(g(x₁)) = 1 + x₁

POL(a) = 0

POL(H(x₁, x₂, x₃)) = x₂

POL(F(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

H(g(a), g(x'''), g(a)) -> F(g(x'''), g(a))

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

H(g(a), g(x'''), g(g(a))) -> F(g(x'''), g(g(a)))
H(g(a), g(x'''), g(g(g(x'''''''')))) -> F(g(x'''), g(g(g(x''''''''))))
F(g(x'), g(g(g(x'''''')))) -> H(g(g(g(x''''''))), x', g(g(g(x''''''))))
H(g(a), g(x'''), g(g(x''''''))) -> F(g(x'''), g(g(x'''''')))
H(g(g(a)), y', z') -> H(g(a), y', z')
H(g(g(g(x''''))), y', z') -> H(g(g(x'''')), y', z')
H(g(g(x'')), y'', z'') -> H(g(x''), y'', z'')
F(g(x'), g(g(x''''))) -> H(g(g(x'''')), x', g(g(x'''')))
H(g(g(x'')), y'', z'') -> F(y'', f(y'', h(x'', y'', z'')))
F(g(x'), g(g(a))) -> H(g(g(a)), x', g(g(a)))

Rules:

f(a, g(y)) -> g(g(y))
f(g(x), a) -> f(x, g(a))
f(g(x), g(y)) -> h(g(y), x, g(y))
h(g(x), y, z) -> f(y, h(x, y, z))
h(a, y, z) -> z

Strategy:

innermost

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

POL(g(x₁))	= 0
POL(h(x₁, x₂, x₃))	= x₃
POL(a)	= 1
POL(H(x₁, x₂, x₃))	= x₃
POL(f(x₁, x₂))	= 0
POL(F(x₁, x₂))	= x₂

POL(g(x₁))	= 1 + x₁
POL(a)	= 0
POL(H(x₁, x₂, x₃))	= x₂
POL(F(x₁, x₂))	= x₁