interpretations
TIMEOUT
We are left with following problem, upon which TcT provides the
certificate TIMEOUT.
Strict Trs:
{ 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())
, f(t, x) -> f'(t, g(x))
, foldC(t, 0()) -> t
, foldC(t, s(n)) -> f(foldC(t, n), C())
, f'(triple(a, b, c), A()) -> f''(foldB(triple(s(a), 0(), c), b))
, f'(triple(a, b, c), B()) -> f(triple(a, b, c), A())
, f'(triple(a, b, c), C()) -> triple(a, b, s(c))
, 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) }
Obligation:
innermost runtime complexity
Answer:
TIMEOUT
Computation stopped due to timeout after 20.0 seconds.
Arrrr..
lmpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ 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) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
mpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ 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) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ 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) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar-ps
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ 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) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..