Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(x, g(y)) -> F(h(x), i(x, y))
F(x, g(y)) -> I(x, y)
I(x, j(y, z)) -> J(g(y), i(x, z))
I(x, j(y, z)) -> I(x, z)
I(h(x), j(j(y, z), 0)) -> J(i(h(x), j(y, z)), i(x, j(y, z)))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
J(g(x), g(y)) -> J(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

J(g(x), g(y)) -> J(x, y)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

J(g(x), g(y)) -> J(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

J(g(x), g(y)) -> J(x, y)

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

g(x₁) -> g(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)

and get the following Size-Change Graph(s):

{1, 2, 3}	,	{1, 2, 3}
1	>	1
2	>	2

{1, 2, 3}	,	{1, 2, 3}
1	=	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1, 2, 3}	,	{1, 2, 3}
1	=	1
2	>	2

{1, 2, 3}	,	{1, 2, 3}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

h(x₁) -> h(x₁)
j(x₁, x₂) -> j(x₁, x₂)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pair:

F(x, g(y)) -> F(h(x), i(x, y))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Strategy:

innermost

As we are in the innermost case, we can delete all 1 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pair:

F(x, g(y)) -> F(h(x), i(x, y))

Rules:

i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Strategy:

innermost

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

F(x, g(y)) -> F(h(x), i(x, y))

three new Dependency Pairs are created:

F(x'', g(j(0, 0))) -> F(h(x''), g(0))
F(x'', g(j(y'', z'))) -> F(h(x''), j(g(y''), i(x'', z')))
F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Instantiation Transformation

Dependency Pairs:

F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))
F(x'', g(j(y'', z'))) -> F(h(x''), j(g(y''), i(x'', z')))

Rules:

i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Strategy:

innermost

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

F(x'', g(j(y'', z'))) -> F(h(x''), j(g(y''), i(x'', z')))

two new Dependency Pairs are created:

F(h(x''''), g(j(y''', z''))) -> F(h(h(x'''')), j(g(y'''), i(h(x''''), z'')))
F(h(h(x'''')), g(j(y''', z''))) -> F(h(h(h(x''''))), j(g(y'''), i(h(h(x'''')), z'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Instantiation Transformation

Dependency Pairs:

F(h(h(x'''')), g(j(y''', z''))) -> F(h(h(h(x''''))), j(g(y'''), i(h(h(x'''')), z'')))
F(h(x''''), g(j(y''', z''))) -> F(h(h(x'''')), j(g(y'''), i(h(x''''), z'')))
F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))

Rules:

i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Strategy:

innermost

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

F(h(x''), g(j(j(y'', z'), 0))) -> F(h(h(x'')), j(i(h(x''), j(y'', z')), i(x'', j(y'', z'))))

three new Dependency Pairs are created:

F(h(h(x'''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''))), j(i(h(h(x'''')), j(y''', z'')), i(h(x''''), j(y''', z''))))
F(h(h(x'''''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''''))), j(i(h(h(x'''''')), j(y''', z'')), i(h(x''''''), j(y''', z''))))
F(h(h(h(x''''''))), g(j(j(y''', z''), 0))) -> F(h(h(h(h(x'''''')))), j(i(h(h(h(x''''''))), j(y''', z'')), i(h(h(x'''''')), j(y''', z''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(h(h(h(x''''''))), g(j(j(y''', z''), 0))) -> F(h(h(h(h(x'''''')))), j(i(h(h(h(x''''''))), j(y''', z'')), i(h(h(x'''''')), j(y''', z''))))
F(h(h(x'''''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''''))), j(i(h(h(x'''''')), j(y''', z'')), i(h(x''''''), j(y''', z''))))
F(h(h(x'''')), g(j(j(y''', z''), 0))) -> F(h(h(h(x''''))), j(i(h(h(x'''')), j(y''', z'')), i(h(x''''), j(y''', z''))))
F(h(x''''), g(j(y''', z''))) -> F(h(h(x'''')), j(g(y'''), i(h(x''''), z'')))
F(h(h(x'''')), g(j(y''', z''))) -> F(h(h(h(x''''))), j(g(y'''), i(h(h(x'''')), z'')))

Rules:

i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Strategy:

innermost

The Proof could not be continued due to a Timeout.
Innermost Termination of R could not be shown.
Duration:
1:00 minutes