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:

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FOLDF2(x, cons2(y, z)) -> FOLDF2(x, z)
FOLDF2(x, cons2(y, z)) -> F2(foldf2(x, z), y)
F'2(triple3(a, b, c), A) -> F''1(foldf2(triple3(cons2(A, a), nil, c), b))
F2(t, x) -> F'2(t, g1(x))
F2(t, x) -> G1(x)
F'2(triple3(a, b, c), A) -> FOLDF2(triple3(cons2(A, a), nil, c), b)
F''1(triple3(a, b, c)) -> FOLDF2(triple3(a, b, nil), c)
F'2(triple3(a, b, c), B) -> F2(triple3(a, b, c), A)

The TRS R consists of the following rules:

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(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:

FOLDF2(x, cons2(y, z)) -> FOLDF2(x, z)
FOLDF2(x, cons2(y, z)) -> F2(foldf2(x, z), y)
F'2(triple3(a, b, c), A) -> F''1(foldf2(triple3(cons2(A, a), nil, c), b))
F2(t, x) -> F'2(t, g1(x))
F2(t, x) -> G1(x)
F'2(triple3(a, b, c), A) -> FOLDF2(triple3(cons2(A, a), nil, c), b)
F''1(triple3(a, b, c)) -> FOLDF2(triple3(a, b, nil), c)
F'2(triple3(a, b, c), B) -> F2(triple3(a, b, c), A)

The TRS R consists of the following rules:

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

FOLDF2(x, cons2(y, z)) -> FOLDF2(x, z)
FOLDF2(x, cons2(y, z)) -> F2(foldf2(x, z), y)
F'2(triple3(a, b, c), A) -> F''1(foldf2(triple3(cons2(A, a), nil, c), b))
F2(t, x) -> F'2(t, g1(x))
F'2(triple3(a, b, c), A) -> FOLDF2(triple3(cons2(A, a), nil, c), b)
F''1(triple3(a, b, c)) -> FOLDF2(triple3(a, b, nil), c)
F'2(triple3(a, b, c), B) -> F2(triple3(a, b, c), A)

The TRS R consists of the following rules:

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


FOLDF2(x, cons2(y, z)) -> FOLDF2(x, z)
FOLDF2(x, cons2(y, z)) -> F2(foldf2(x, z), y)
The remaining pairs can at least be oriented weakly.

F'2(triple3(a, b, c), A) -> F''1(foldf2(triple3(cons2(A, a), nil, c), b))
F2(t, x) -> F'2(t, g1(x))
F'2(triple3(a, b, c), A) -> FOLDF2(triple3(cons2(A, a), nil, c), b)
F''1(triple3(a, b, c)) -> FOLDF2(triple3(a, b, nil), c)
F'2(triple3(a, b, c), B) -> F2(triple3(a, b, c), A)
Used ordering: Polynomial interpretation [21]:

POL(A) = 0   
POL(B) = 0   
POL(C) = 0   
POL(F2(x1, x2)) = 2·x1   
POL(F'2(x1, x2)) = 2·x1   
POL(F''1(x1)) = 2·x1   
POL(FOLDF2(x1, x2)) = 2·x1 + 2·x2   
POL(cons2(x1, x2)) = 2 + x2   
POL(f2(x1, x2)) = 2 + x1   
POL(f'2(x1, x2)) = 2 + x1   
POL(f''1(x1)) = 1 + x1   
POL(foldf2(x1, x2)) = x1 + x2   
POL(g1(x1)) = 0   
POL(nil) = 0   
POL(triple3(x1, x2, x3)) = x2 + x3   

The following usable rules [14] were oriented:

f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
foldf2(x, nil) -> x
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f2(t, x) -> f'2(t, g1(x))
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))



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

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

F2(t, x) -> F'2(t, g1(x))
F'2(triple3(a, b, c), A) -> F''1(foldf2(triple3(cons2(A, a), nil, c), b))
F'2(triple3(a, b, c), A) -> FOLDF2(triple3(cons2(A, a), nil, c), b)
F'2(triple3(a, b, c), B) -> F2(triple3(a, b, c), A)
F''1(triple3(a, b, c)) -> FOLDF2(triple3(a, b, nil), c)

The TRS R consists of the following rules:

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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:

F2(t, x) -> F'2(t, g1(x))
F'2(triple3(a, b, c), B) -> F2(triple3(a, b, c), A)

The TRS R consists of the following rules:

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(t, x) -> F'2(t, g1(x))
F'2(triple3(a, b, c), B) -> F2(triple3(a, b, c), A)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(A) = 0   
POL(B) = 2   
POL(C) = 2   
POL(F2(x1, x2)) = 1 + 3·x2   
POL(F'2(x1, x2)) = 2·x2   
POL(g1(x1)) = x1   
POL(triple3(x1, x2, x3)) = 0   

The following usable rules [14] were oriented:

g1(C) -> A
g1(B) -> A
g1(C) -> B
g1(A) -> A
g1(B) -> B
g1(C) -> C



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

g1(A) -> A
g1(B) -> A
g1(B) -> B
g1(C) -> A
g1(C) -> B
g1(C) -> C
foldf2(x, nil) -> x
foldf2(x, cons2(y, z)) -> f2(foldf2(x, z), y)
f2(t, x) -> f'2(t, g1(x))
f'2(triple3(a, b, c), C) -> triple3(a, b, cons2(C, c))
f'2(triple3(a, b, c), B) -> f2(triple3(a, b, c), A)
f'2(triple3(a, b, c), A) -> f''1(foldf2(triple3(cons2(A, a), nil, c), b))
f''1(triple3(a, b, c)) -> foldf2(triple3(a, b, nil), c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.