Termination Proof using AProVE

Term Rewriting System R:
[z, x, y]
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

H(z, e(x)) -> H(c(z), d(z, x))
H(z, e(x)) -> D(z, x)
D(z, g(x, y)) -> G(e(x), d(z, y))
D(z, g(x, y)) -> D(z, y)
D(c(z), g(g(x, y), 0)) -> G(d(c(z), g(x, y)), d(z, g(x, y)))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
G(e(x), e(y)) -> G(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

G(e(x), e(y)) -> G(x, y)

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(e(x), e(y)) -> G(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, x₂)) = x₁

POL(e(x₁)) = 1 + 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:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(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:

D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(z, g(x, y)) -> D(z, y)

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

D(c(z), g(g(x, y), 0)) -> D(z, g(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(c(x₁)) = 1 + x₁

POL(0) = 0

POL(g(x₁, x₂)) = 0

POL(e(x₁)) = 0

POL(D(x₁, x₂)) = x₁

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:

D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(z, g(x, y)) -> D(z, y)

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = 0

POL(0) = 1

POL(g(x₁, x₂)) = x₁ + x₂

POL(e(x₁)) = 0

POL(D(x₁, x₂)) = 1 + 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:

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

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

D(z, g(x, y)) -> D(z, 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, x₂)) = 1 + x₂

POL(D(x₁, x₂)) = 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:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(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:

H(z, e(x)) -> H(c(z), d(z, x))

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

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

H(z, e(x)) -> H(c(z), d(z, x))

three new Dependency Pairs are created:

H(z'', e(g(0, 0))) -> H(c(z''), e(0))
H(z'', e(g(x'', y'))) -> H(c(z''), g(e(x''), d(z'', y')))
H(c(z''), e(g(g(x'', y'), 0))) -> H(c(c(z'')), g(d(c(z''), g(x'', y')), d(z'', g(x'', y'))))

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:

H(c(z''), e(g(g(x'', y'), 0))) -> H(c(c(z'')), g(d(c(z''), g(x'', y')), d(z'', g(x'', y'))))
H(z'', e(g(x'', y'))) -> H(c(z''), g(e(x''), d(z'', y')))

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

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

H(z'', e(g(x'', y'))) -> H(c(z''), g(e(x''), d(z'', y')))

two new Dependency Pairs are created:

H(c(z''''), e(g(x''', y''))) -> H(c(c(z'''')), g(e(x'''), d(c(z''''), y'')))
H(c(c(z'''')), e(g(x''', y''))) -> H(c(c(c(z''''))), g(e(x'''), d(c(c(z'''')), y'')))

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:

H(c(c(z'''')), e(g(x''', y''))) -> H(c(c(c(z''''))), g(e(x'''), d(c(c(z'''')), y'')))
H(c(z''''), e(g(x''', y''))) -> H(c(c(z'''')), g(e(x'''), d(c(z''''), y'')))
H(c(z''), e(g(g(x'', y'), 0))) -> H(c(c(z'')), g(d(c(z''), g(x'', y')), d(z'', g(x'', y'))))

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

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

H(c(z''), e(g(g(x'', y'), 0))) -> H(c(c(z'')), g(d(c(z''), g(x'', y')), d(z'', g(x'', y'))))

three new Dependency Pairs are created:

H(c(c(z'''')), e(g(g(x''', y''), 0))) -> H(c(c(c(z''''))), g(d(c(c(z'''')), g(x''', y'')), d(c(z''''), g(x''', y''))))
H(c(c(z'''''')), e(g(g(x''', y''), 0))) -> H(c(c(c(z''''''))), g(d(c(c(z'''''')), g(x''', y'')), d(c(z''''''), g(x''', y''))))
H(c(c(c(z''''''))), e(g(g(x''', y''), 0))) -> H(c(c(c(c(z'''''')))), g(d(c(c(c(z''''''))), g(x''', y'')), d(c(c(z'''''')), g(x''', y''))))

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:

H(c(c(c(z''''''))), e(g(g(x''', y''), 0))) -> H(c(c(c(c(z'''''')))), g(d(c(c(c(z''''''))), g(x''', y'')), d(c(c(z'''''')), g(x''', y''))))
H(c(c(z'''''')), e(g(g(x''', y''), 0))) -> H(c(c(c(z''''''))), g(d(c(c(z'''''')), g(x''', y'')), d(c(z''''''), g(x''', y''))))
H(c(c(z'''')), e(g(g(x''', y''), 0))) -> H(c(c(c(z''''))), g(d(c(c(z'''')), g(x''', y'')), d(c(z''''), g(x''', y''))))
H(c(z''''), e(g(x''', y''))) -> H(c(c(z'''')), g(e(x'''), d(c(z''''), y'')))
H(c(c(z'''')), e(g(x''', y''))) -> H(c(c(c(z''''))), g(e(x'''), d(c(c(z'''')), y'')))

Rules:

h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))

Strategy:

innermost

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

POL(c(x₁))	= 1 + x₁
POL(0)	= 0
POL(g(x₁, x₂))	= 0
POL(e(x₁))	= 0
POL(D(x₁, x₂))	= x₁

POL(g(x₁, x₂))	= 1 + x₂
POL(D(x₁, x₂))	= x₂