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)

Innermost 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
         ↳Argument Filtering and Ordering


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)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


The following usable rules for innermost w.r.t. to the 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 + x2  
  POL(FOLDC(x1, x2))=  x1 + x2  
  POL(triple(x1, x2))=  x1 + x2  
  POL(f'(x1, x2))=  1 + x1 + x2  
  POL(f(x1, x2))=  1 + x1 + x2  
  POL(F(x1, x2))=  x1 + x2  
  POL(foldC(x1, x2))=  x1 + x2  
  POL(FOLDB(x1, x2))=  x1 + x2  
  POL(F''(x1))=  x1  
  POL(C)=  0  
  POL(0)=  0  
  POL(B)=  0  
  POL(g(x1))=  x1  
  POL(F'(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  
  POL(f''(x1))=  x1  
  POL(A)=  0  

resulting in one new DP problem.
Used Argument Filtering System:
FOLDC(x1, x2) -> FOLDC(x1, x2)
F(x1, x2) -> F(x1, x2)
s(x1) -> s(x1)
foldC(x1, x2) -> foldC(x1, x2)
FOLDB(x1, x2) -> FOLDB(x1, x2)
foldB(x1, x2) -> foldB(x1, x2)
F'(x1, x2) -> F'(x1, x2)
g(x1) -> g(x1)
triple(x1, x2, x3) -> triple(x2, x3)
F''(x1) -> F''(x1)
f(x1, x2) -> f(x1, x2)
f'(x1, x2) -> f'(x1, x2)
f''(x1) -> f''(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

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


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)


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Narrowing Transformation


Dependency Pairs:

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


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 no new DP problems.


Innermost Termination of R successfully shown.
Duration:
0:04 minutes