Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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
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)
fold(t, x, 0) → t
fold(t, x, s(n)) → f(fold(t, x, n), x)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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)
FOLD(t, x, s(n)) → F(fold(t, x, n), x)
FOLD(t, x, s(n)) → FOLD(t, x, n)

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
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)
fold(t, x, 0) → t
fold(t, x, s(n)) → f(fold(t, x, n), x)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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)
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)
FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDC(t, s(n)) → FOLDC(t, n)
F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)
FOLDB(t, s(n)) → F(foldB(t, n), B)
FOLDB(t, s(n)) → FOLDB(t, n)

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
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)
fold(t, x, 0) → t
fold(t, x, s(n)) → f(fold(t, x, n), x)

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

(6) Obligation:

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

FOLD(t, x, s(n)) → FOLD(t, x, n)

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
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)
fold(t, x, 0) → t
fold(t, x, s(n)) → f(fold(t, x, n), x)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FOLD(t, x, s(n)) → FOLD(t, x, n)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FOLD(x1, x2, x3)  =  FOLD(x1, x2, x3)
s(x1)  =  s(x1)
g(x1)  =  g(x1)
A  =  A
B  =  B
C  =  C
foldB(x1, x2)  =  x1
0  =  0
f(x1, x2)  =  f
foldC(x1, x2)  =  x1
f'(x1, x2)  =  f'
triple(x1, x2, x3)  =  triple
f''(x1)  =  x1
fold(x1, x2, x3)  =  x1

Recursive Path Order [RPO].
Precedence:

g₁ > A > [FOLD₃, s₁, f, f', triple]
g₁ > B > [FOLD₃, s₁, f, f', triple]
C > A > [FOLD₃, s₁, f, f', triple]
C > B > [FOLD₃, s₁, f, f', triple]
0 > [FOLD₃, s₁, f, f', triple]

The following usable rules [FROCOS05] were oriented:

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)
fold(t, x, 0) → t
fold(t, x, s(n)) → f(fold(t, x, n), x)

(8) Obligation:

Q DP problem:
P is empty.
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
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)
fold(t, x, 0) → t
fold(t, x, s(n)) → f(fold(t, x, n), x)

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

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE