Termination Proof using AProVE

Term Rewriting System R:
[x, y, z, t, a, b, c]
g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FOLDF(x, cons(y, z)) -> F(foldf(x, z), y)
FOLDF(x, cons(y, z)) -> FOLDF(x, z)
F(t, x) -> F'(t, g(x))
F(t, x) -> G(x)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F'(triple(a, b, c), A) -> F''(foldf(triple(cons(A, a), nil, c), b))
F'(triple(a, b, c), A) -> FOLDF(triple(cons(A, a), nil, c), b)
F''(triple(a, b, c)) -> FOLDF(triple(a, b, nil), c)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

Dependency Pairs:

F'(triple(a, b, c), A) -> FOLDF(triple(cons(A, a), nil, c), b)
FOLDF(x, cons(y, z)) -> FOLDF(x, z)
F''(triple(a, b, c)) -> FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), A) -> F''(foldf(triple(cons(A, a), nil, c), b))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDF(x, cons(y, z)) -> F(foldf(x, z), y)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)

The following dependency pairs can be strictly oriented:

F'(triple(a, b, c), A) -> FOLDF(triple(cons(A, a), nil, c), b)
F'(triple(a, b, c), A) -> F''(foldf(triple(cons(A, a), nil, c), b))

Additionally, the following usable rules using the Ce-refinement can be oriented:

foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f(t, x) -> f'(t, g(x))
g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(triple(x₁, x₂, x₃)) = x₂ + x₃

POL(f'(x₁, x₂)) = x₁ + x₂

POL(F(x₁, x₂)) = x₁ + x₂

POL(f(x₁, x₂)) = x₁ + x₂

POL(FOLDF(x₁, x₂)) = x₁ + x₂

POL(F''(x₁)) = x₁

POL(C) = 1

POL(B) = 1

POL(g(x₁)) = x₁

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nil) = 0

POL(F'(x₁, x₂)) = x₁ + x₂

POL(foldf(x₁, x₂)) = x₁ + x₂

POL(f''(x₁)) = x₁

POL(A) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

FOLDF(x, cons(y, z)) -> FOLDF(x, z)
F''(triple(a, b, c)) -> FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDF(x, cons(y, z)) -> F(foldf(x, z), y)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)

Using the Dependency Graph the DP problem was split into 2 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 3

                 ↳Polynomial Ordering

Dependency Pairs:

F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)

The following dependency pair can be strictly oriented:

F'(triple(a, b, c), B) -> F(triple(a, b, c), A)

Additionally, the following usable rules using the Ce-refinement can be oriented:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(C) = 1

POL(g(x₁)) = x₁

POL(B) = 1

POL(F'(x₁, x₂)) = x₂

POL(triple(x₁, x₂, x₃)) = 0

POL(A) = 0

POL(F(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 5

                 ↳Dependency Graph

Dependency Pair:

F(t, x) -> F'(t, g(x))

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pair:

FOLDF(x, cons(y, z)) -> FOLDF(x, z)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)

The following dependency pair can be strictly oriented:

FOLDF(x, cons(y, z)) -> FOLDF(x, z)

There are no usable rules using the Ce-refinement that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = 1 + x₂

POL(FOLDF(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldf(x, nil) -> x
foldf(x, cons(y, z)) -> f(foldf(x, z), y)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, cons(C, c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) -> foldf(triple(a, b, nil), c)

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

POL(triple(x₁, x₂, x₃))	= x₂ + x₃
POL(f'(x₁, x₂))	= x₁ + x₂
POL(F(x₁, x₂))	= x₁ + x₂
POL(f(x₁, x₂))	= x₁ + x₂
POL(FOLDF(x₁, x₂))	= x₁ + x₂
POL(F''(x₁))	= x₁
POL(C)	= 1
POL(B)	= 1
POL(g(x₁))	= x₁
POL(cons(x₁, x₂))	= x₁ + x₂
POL(nil)	= 0
POL(F'(x₁, x₂))	= x₁ + x₂
POL(foldf(x₁, x₂))	= x₁ + x₂
POL(f''(x₁))	= x₁
POL(A)	= 1

POL(cons(x₁, x₂))	= 1 + x₂
POL(FOLDF(x₁, x₂))	= x₂