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:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F2(x, s1(y)) -> -12(s1(y), x)
F2(x, s1(y)) -> -12(x, s1(y))
F2(s1(x), y) -> P1(-2(y, s1(x)))
F2(s1(x), y) -> P1(-2(s1(x), y))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), y) -> F2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
F2(x, s1(y)) -> F2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
F2(x, s1(y)) -> P1(-2(s1(y), x))
F2(x, s1(y)) -> P1(-2(x, s1(y)))
F2(s1(x), y) -> -12(y, s1(x))
F2(s1(x), y) -> -12(s1(x), y)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
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:
F2(x, s1(y)) -> -12(s1(y), x)
F2(x, s1(y)) -> -12(x, s1(y))
F2(s1(x), y) -> P1(-2(y, s1(x)))
F2(s1(x), y) -> P1(-2(s1(x), y))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), y) -> F2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
F2(x, s1(y)) -> F2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
F2(x, s1(y)) -> P1(-2(s1(y), x))
F2(x, s1(y)) -> P1(-2(x, s1(y)))
F2(s1(x), y) -> -12(y, s1(x))
F2(s1(x), y) -> -12(s1(x), y)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 8 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-12(s1(x), s1(y)) -> -12(x, y)
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
-12(s1(x), s1(y)) -> -12(x, y)
Used argument filtering: -12(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(s1(x), y) -> F2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
F2(x, s1(y)) -> F2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
The TRS R consists of the following rules:
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), y) -> f2(p1(-2(s1(x), y)), p1(-2(y, s1(x))))
f2(x, s1(y)) -> f2(p1(-2(x, s1(y))), p1(-2(s1(y), x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.