Termination Proof using AProVE

Term Rewriting System R:
[t, n, x, a, b, c]
g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FOLDB(t, s(n)) -> F(foldB(t, n), B)
FOLDB(t, s(n)) -> FOLDB(t, n)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
FOLDC(t, s(n)) -> FOLDC(t, n)
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''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

FOLDB(t, s(n)) -> FOLDB(t, n)
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
FOLDC(t, s(n)) -> FOLDC(t, n)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, x) -> F'(t, g(x))
FOLDB(t, s(n)) -> F(foldB(t, n), B)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

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

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

six new Dependency Pairs are created:

F(t, A) -> F'(t, A)
F(t, B) -> F'(t, A)
F(t, B) -> F'(t, B)
F(t, C) -> F'(t, A)
F(t, C) -> F'(t, B)
F(t, C) -> F'(t, C)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

F(t, B) -> F'(t, B)
FOLDC(t, s(n)) -> FOLDC(t, n)
F(t, A) -> F'(t, A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, C) -> F'(t, B)
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F(t, C) -> F'(t, A)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))
F(t, B) -> F'(t, A)
FOLDB(t, s(n)) -> F(foldB(t, n), B)
FOLDB(t, s(n)) -> FOLDB(t, n)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

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

F'(triple(a, b, c), A) -> F''(foldB(triple(s(a), 0, c), b))

two new Dependency Pairs are created:

F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F'(triple(a', s(n'), c'), A) -> F''(f(foldB(triple(s(a'), 0, c'), n'), B))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

FOLDB(t, s(n)) -> FOLDB(t, n)
FOLDC(t, s(n)) -> FOLDC(t, n)
F(t, C) -> F'(t, B)
F'(triple(a', s(n'), c'), A) -> F''(f(foldB(triple(s(a'), 0, c'), n'), B))
F(t, C) -> F'(t, A)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F(t, B) -> F'(t, A)
FOLDB(t, s(n)) -> F(foldB(t, n), B)
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F(t, A) -> F'(t, A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, B) -> F'(t, B)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

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

F'(triple(a', s(n'), c'), A) -> F''(f(foldB(triple(s(a'), 0, c'), n'), B))

three new Dependency Pairs are created:

F'(triple(a'', s(n''), c''), A) -> F''(f'(foldB(triple(s(a''), 0, c''), n''), g(B)))
F'(triple(a'', s(0), c''), A) -> F''(f(triple(s(a''), 0, c''), B))
F'(triple(a'', s(s(n'')), c''), A) -> F''(f(f(foldB(triple(s(a''), 0, c''), n''), B), B))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Instantiation Transformation

Dependency Pairs:

F(t, B) -> F'(t, B)
FOLDC(t, s(n)) -> FOLDC(t, n)
F(t, A) -> F'(t, A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, C) -> F'(t, B)
F'(triple(a'', s(s(n'')), c''), A) -> F''(f(f(foldB(triple(s(a''), 0, c''), n''), B), B))
F'(triple(a'', s(0), c''), A) -> F''(f(triple(s(a''), 0, c''), B))
F'(triple(a'', s(n''), c''), A) -> F''(f'(foldB(triple(s(a''), 0, c''), n''), g(B)))
F(t, C) -> F'(t, A)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F(t, B) -> F'(t, A)
FOLDB(t, s(n)) -> F(foldB(t, n), B)
FOLDB(t, s(n)) -> FOLDB(t, n)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

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

F(t, A) -> F'(t, A)

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

FOLDB(t, s(n)) -> FOLDB(t, n)
FOLDC(t, s(n)) -> FOLDC(t, n)
F(t, C) -> F'(t, B)
F'(triple(a'', s(s(n'')), c''), A) -> F''(f(f(foldB(triple(s(a''), 0, c''), n''), B), B))
F'(triple(a'', s(0), c''), A) -> F''(f(triple(s(a''), 0, c''), B))
F'(triple(a'', s(n''), c''), A) -> F''(f'(foldB(triple(s(a''), 0, c''), n''), g(B)))
F(t, C) -> F'(t, A)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F(t, B) -> F'(t, A)
FOLDB(t, s(n)) -> F(foldB(t, n), B)
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F(triple(a'', b'', c''), A) -> F'(triple(a'', b'', c''), A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, B) -> F'(t, B)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

The following dependency pairs can be strictly oriented:

F'(triple(a'', s(s(n'')), c''), A) -> F''(f(f(foldB(triple(s(a''), 0, c''), n''), B), B))
F'(triple(a'', s(0), c''), A) -> F''(f(triple(s(a''), 0, c''), B))
F'(triple(a'', s(n''), c''), A) -> F''(f'(foldB(triple(s(a''), 0, c''), n''), g(B)))

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

foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(foldB(x₁, x₂)) = x₁

POL(FOLDC(x₁, x₂)) = x₁

POL(triple(x₁, x₂, x₃)) = x₂

POL(f'(x₁, x₂)) = x₁

POL(F(x₁, x₂)) = x₁

POL(f(x₁, x₂)) = x₁

POL(foldC(x₁, x₂)) = x₁

POL(FOLDB(x₁, x₂)) = x₁

POL(F''(x₁)) = x₁

POL(0) = 0

POL(C) = 0

POL(B) = 0

POL(g(x₁)) = 0

POL(F'(x₁, x₂)) = x₁

POL(s(x₁)) = 1

POL(f''(x₁)) = x₁

POL(A) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Instantiation Transformation

Dependency Pairs:

FOLDB(t, s(n)) -> FOLDB(t, n)
FOLDC(t, s(n)) -> FOLDC(t, n)
F(t, C) -> F'(t, B)
F(t, C) -> F'(t, A)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F(t, B) -> F'(t, A)
FOLDB(t, s(n)) -> F(foldB(t, n), B)
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F(triple(a'', b'', c''), A) -> F'(triple(a'', b'', c''), A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, B) -> F'(t, B)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

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

F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)

one new Dependency Pair is created:

F''(triple(s(a'''), 0, c')) -> FOLDC(triple(s(a'''), 0, 0), c')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

F(t, B) -> F'(t, B)
FOLDC(t, s(n)) -> FOLDC(t, n)
F(triple(a'', b'', c''), A) -> F'(triple(a'', b'', c''), A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
F(t, C) -> F'(t, B)
F(t, C) -> F'(t, A)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(s(a'''), 0, c')) -> FOLDC(triple(s(a'''), 0, 0), c')
F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
F(t, B) -> F'(t, A)
FOLDB(t, s(n)) -> F(foldB(t, n), B)
FOLDB(t, s(n)) -> FOLDB(t, n)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

The following dependency pairs can be strictly oriented:

FOLDB(t, s(n)) -> F(foldB(t, n), B)
FOLDB(t, s(n)) -> FOLDB(t, n)

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

foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
f(t, x) -> f'(t, g(x))
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(foldB(x₁, x₂)) = x₁

POL(FOLDC(x₁, x₂)) = x₁

POL(triple(x₁, x₂, x₃)) = x₂

POL(f'(x₁, x₂)) = x₁

POL(f(x₁, x₂)) = x₁

POL(foldC(x₁, x₂)) = x₁

POL(F(x₁, x₂)) = x₁

POL(FOLDB(x₁, x₂)) = x₁ + x₂

POL(F''(x₁)) = 0

POL(0) = 0

POL(C) = 0

POL(B) = 0

POL(g(x₁)) = 0

POL(F'(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(f''(x₁)) = x₁

POL(A) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pairs:

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 9

                 ↳Polynomial Ordering

Dependency Pairs:

FOLDC(t, s(n)) -> FOLDC(t, n)
F(t, C) -> F'(t, B)
F(t, C) -> F'(t, A)
FOLDC(t, s(n)) -> F(foldC(t, n), C)
F''(triple(s(a'''), 0, c')) -> FOLDC(triple(s(a'''), 0, 0), c')
F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F(triple(a'', b'', c''), A) -> F'(triple(a'', b'', c''), A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

The following dependency pairs can be strictly oriented:

FOLDC(t, s(n)) -> FOLDC(t, n)
FOLDC(t, s(n)) -> F(foldC(t, n), C)

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

foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)
f(t, x) -> f'(t, g(x))
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
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(foldB(x₁, x₂)) = x₁

POL(FOLDC(x₁, x₂)) = x₁ + x₂

POL(triple(x₁, x₂, x₃)) = x₃

POL(f'(x₁, x₂)) = x₁ + x₂

POL(F(x₁, x₂)) = x₁

POL(f(x₁, x₂)) = x₁ + x₂

POL(foldC(x₁, x₂)) = x₁ + x₂

POL(F''(x₁)) = x₁

POL(0) = 0

POL(C) = 1

POL(B) = 0

POL(g(x₁)) = x₁

POL(F'(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(f''(x₁)) = x₁

POL(A) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 10

                 ↳Dependency Graph

Dependency Pairs:

F(t, C) -> F'(t, B)
F(t, C) -> F'(t, A)
F''(triple(s(a'''), 0, c')) -> FOLDC(triple(s(a'''), 0, 0), c')
F'(triple(a', 0, c'), A) -> F''(triple(s(a'), 0, c'))
F(triple(a'', b'', c''), A) -> F'(triple(a'', b'', c''), A)
F'(triple(a, b, c), B) -> F(triple(a, b, c), A)

Rules:

g(A) -> A
g(B) -> A
g(B) -> B
g(C) -> A
g(C) -> B
g(C) -> C
foldB(t, 0) -> t
foldB(t, s(n)) -> f(foldB(t, n), B)
foldC(t, 0) -> t
foldC(t, s(n)) -> f(foldC(t, n), C)
f(t, x) -> f'(t, g(x))
f'(triple(a, b, c), C) -> triple(a, b, s(c))
f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes

POL(foldB(x₁, x₂))	= x₁
POL(FOLDC(x₁, x₂))	= x₁
POL(triple(x₁, x₂, x₃))	= x₂
POL(f'(x₁, x₂))	= x₁
POL(F(x₁, x₂))	= x₁
POL(f(x₁, x₂))	= x₁
POL(foldC(x₁, x₂))	= x₁
POL(FOLDB(x₁, x₂))	= x₁
POL(F''(x₁))	= x₁
POL(0)	= 0
POL(C)	= 0
POL(B)	= 0
POL(g(x₁))	= 0
POL(F'(x₁, x₂))	= x₁
POL(s(x₁))	= 1
POL(f''(x₁))	= x₁
POL(A)	= 0