Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following 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)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following 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)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FOLDF(x, cons(y, z)) → FOLDF(x, z)
F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
F(t, x) → G(x)
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F(t, x) → F'(t, g(x))
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

FOLDF(x, cons(y, z)) → FOLDF(x, z)
F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
F(t, x) → G(x)
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F(t, x) → F'(t, g(x))
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

FOLDF(x, cons(y, z)) → FOLDF(x, z)
F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F(t, x) → F'(t, g(x))
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


FOLDF(x, cons(y, z)) → FOLDF(x, z)
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
The remaining pairs can at least be oriented weakly.

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))
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)
Used ordering: Polynomial interpretation [25,35]:

POL(F''(x1)) = (2)x_1   
POL(f''(x1)) = x_1   
POL(g(x1)) = 0   
POL(f(x1, x2)) = 1 + x_1   
POL(C) = 0   
POL(FOLDF(x1, x2)) = (2)x_1 + (2)x_2   
POL(A) = 0   
POL(triple(x1, x2, x3)) = (2)x_2 + x_3   
POL(cons(x1, x2)) = 1 + x_2   
POL(f'(x1, x2)) = 1 + x_1   
POL(F'(x1, x2)) = (2)x_1   
POL(B) = 0   
POL(foldf(x1, x2)) = x_1 + x_2   
POL(F(x1, x2)) = (2)x_1   
POL(nil) = 0   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

foldf(x, nil) → x
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
F(t, x) → F'(t, g(x))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F(t, x) → F'(t, g(x))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(t, x) → F'(t, g(x))
The remaining pairs can at least be oriented weakly.

F'(triple(a, b, c), B) → F(triple(a, b, c), A)
Used ordering: Polynomial interpretation [25,35]:

POL(F'(x1, x2)) = 3 + x_1 + (3)x_2   
POL(g(x1)) = x_1   
POL(B) = 4   
POL(C) = 4   
POL(A) = 2   
POL(F(x1, x2)) = 4 + (4)x_1 + (4)x_2   
POL(triple(x1, x2, x3)) = 1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F'(triple(a, b, c), B) → F(triple(a, b, c), A)

The TRS R consists of the following 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.