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))

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

         ↳Remaining

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))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = 0

POL(0) = 0

POL(g(x₁, x₂)) = x₂

POL(e(x₁)) = 1 + x₁

POL(G(x₁, x₂)) = x₁

POL(d(x₁, x₂)) = 1

POL(h(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining

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))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Remaining

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))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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))

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)) = 0

POL(h(x₁, x₂)) = 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

         ↳Remaining

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))

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 rules can be oriented:

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))

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)) = 0

POL(h(x₁, x₂)) = 1

POL(D(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

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))
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))

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

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))
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))

Termination of R could not be shown.
Duration:
0:00 minutes

POL(c(x₁))	= 0
POL(0)	= 0
POL(g(x₁, x₂))	= x₂
POL(e(x₁))	= 1 + x₁
POL(G(x₁, x₂))	= x₁
POL(d(x₁, x₂))	= 1
POL(h(x₁, x₂))	= 0