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.



   R
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.


   R
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:



   R
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:



   R
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:



   R
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:



   R
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.



   R
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:



   R
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.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Nar
             ...
               →DP Problem 8
Dependency Graph


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)


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.


   R
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.



   R
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