Termination w.r.t. Q of the following Term Rewriting System could not be shown:
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.
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 contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
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.