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. 



   R
     ↳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. 


   R
     ↳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:

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


Additionally, the following rules can be oriented:

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)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(triple(x1, x2, x3))=  x2 + x3  
  POL(f'(x1, x2))=  1 + x1  
  POL(F(x1, x2))=  x1  
  POL(f(x1, x2))=  1 + x1  
  POL(FOLDF(x1, x2))=  x1 + x2  
  POL(F''(x1))=  x1  
  POL(C)=  0  
  POL(g(x1))=  0  
  POL(B)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(nil)=  0  
  POL(F'(x1, x2))=  x1  
  POL(foldf(x1, x2))=  x1 + x2  
  POL(f''(x1))=  x1  
  POL(A)=  0  

resulting in one new DP problem. 



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


Dependency Pairs:

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)
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))


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 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


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
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 rules can be oriented:

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)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(triple(x1, x2, x3))=  0  
  POL(f'(x1, x2))=  0  
  POL(f(x1, x2))=  0  
  POL(F(x1, x2))=  x2  
  POL(C)=  1  
  POL(g(x1))=  x1  
  POL(B)=  1  
  POL(cons(x1, x2))=  0  
  POL(nil)=  0  
  POL(F'(x1, x2))=  x2  
  POL(foldf(x1, x2))=  x1  
  POL(f''(x1))=  0  
  POL(A)=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳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.

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