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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Innermost 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)
FOLD(t, x, s(n)) -> F(fold(t, x, n), x)
FOLD(t, x, s(n)) -> FOLD(t, x, n)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

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 3

             ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

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 3

             ↳Nar

...

               →DP Problem 4

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

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 3

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

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'(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'(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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

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

FOLDB(t, s(n)) -> FOLDB(t, n)

one new Dependency Pair is created:

FOLDB(t'', s(s(n''))) -> FOLDB(t'', s(n''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

FOLDB(t'', s(s(n''))) -> FOLDB(t'', s(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(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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

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

FOLDC(t, s(n)) -> FOLDC(t, n)

one new Dependency Pair is created:

FOLDC(t'', s(s(n''))) -> FOLDC(t'', s(n''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

F(t, B) -> F'(t, B)
FOLDC(t'', s(s(n''))) -> FOLDC(t'', s(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'(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'(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(s(n''))) -> FOLDB(t'', s(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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

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

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

two new Dependency Pairs are created:

F'(triple(a', s(n''), c'), A) -> FOLDB(triple(s(a'), 0, c'), s(n''))
F'(triple(a', s(s(n'''')), c'), A) -> FOLDB(triple(s(a'), 0, c'), s(s(n'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 11

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FOLDB(t'', s(0)) -> F(t'', B)
FOLDB(t'', s(s(n''))) -> F(f(foldB(t'', n''), B), B)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 12

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

four new Dependency Pairs are created:

F(triple(a''', 0, c'''), C) -> F'(triple(a''', 0, c'''), A)
F(triple(a''', s(n'''), c'''), C) -> F'(triple(a''', s(n'''), c'''), A)
F(triple(a''', s(n''''), c'''), C) -> F'(triple(a''', s(n''''), c'''), A)
F(triple(a''', s(s(n'''''')), c'''), C) -> F'(triple(a''', s(s(n'''''')), c'''), A)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 14

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FOLDC(t'', s(0)) -> F(t'', C)
FOLDC(t'', s(s(n''))) -> F(f(foldC(t'', n''), C), C)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

F'(triple(a''', 0, s(n'''')), A) -> F''(triple(s(a'''), 0, s(n'''')))
F'(triple(a''', 0, s(s(n''''''))), A) -> F''(triple(s(a'''), 0, s(s(n''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

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

five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

FOLDB(t'', s(0)) -> F(t'', B)

five new Dependency Pairs are created:

FOLDB(triple(a''''', 0, c'''''), s(0)) -> F(triple(a''''', 0, c'''''), B)
FOLDB(triple(a''''', s(n'''''), c'''''), s(0)) -> F(triple(a''''', s(n'''''), c'''''), B)
FOLDB(triple(a''''', s(n''''''), c'''''), s(0)) -> F(triple(a''''', s(n''''''), c'''''), B)
FOLDB(triple(a''''', s(s(n'''''''')), c'''''), s(0)) -> F(triple(a''''', s(s(n'''''''')), c'''''), B)
FOLDB(triple(a'''', b'''', c''''), s(0)) -> F(triple(a'''', b'''', c''''), B)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 19

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

FOLDC(t'', s(0)) -> F(t'', C)

five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

Strategy:
innermost
Dependency Pair:
FOLD(t, x, s(n)) -> FOLD(t, x, 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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

Strategy:
innermost
Dependency Pair:
FOLD(t, x, s(n)) -> FOLD(t, x, 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)
fold(t, x, 0) -> t
fold(t, x, s(n)) -> f(fold(t, x, n), x)

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:14 minutes