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

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

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

The following dependency pair can be strictly oriented:

J(g(x), g(y)) -> J(x, 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(g(x₁)) = 1 + x₁

POL(J(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

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

The following dependency pair can be strictly oriented:

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

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(I(x₁, x₂)) = x₁

POL(0) = 0

POL(g(x₁)) = 0

POL(h(x₁)) = 1 + x₁

POL(j(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pairs:

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

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 1

POL(g(x₁)) = 0

POL(h(x₁)) = 0

POL(j(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Polo

...

               →DP Problem 6

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

The following dependency pair can be strictly oriented:

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

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(I(x₁, x₂)) = x₂

POL(j(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Polo

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Narrowing Transformation

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

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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 8

             ↳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:

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

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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 8

             ↳Inst

...

               →DP Problem 9

                 ↳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:

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

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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 8

             ↳Inst

...

               →DP Problem 10

                 ↳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:

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

Innermost Termination of R could not be shown.
Duration:
0:03 minutes

POL(I(x₁, x₂))	= x₁
POL(0)	= 0
POL(g(x₁))	= 0
POL(h(x₁))	= 1 + x₁
POL(j(x₁, x₂))	= 0

POL(I(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 1
POL(g(x₁))	= 0
POL(h(x₁))	= 0
POL(j(x₁, x₂))	= x₁ + x₂

POL(I(x₁, x₂))	= x₂
POL(j(x₁, x₂))	= 1 + x₂