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:
f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z
The set Q consists of the following terms:
f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)
Q DP problem:
The TRS P consists of the following rules:
H3(g1(x), y, z) -> H3(x, y, z)
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
F2(g1(x), a) -> F2(x, g1(a))
The TRS R consists of the following rules:
f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z
The set Q consists of the following terms:
f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
H3(g1(x), y, z) -> H3(x, y, z)
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
F2(g1(x), a) -> F2(x, g1(a))
The TRS R consists of the following rules:
f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z
The set Q consists of the following terms:
f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)
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.
F2(g1(x), a) -> F2(x, g1(a))
Used argument filtering: H3(x1, x2, x3) = x3
F2(x1, x2) = x2
h3(x1, x2, x3) = x3
g1(x1) = g
a = a
f2(x1, x2) = f
Used ordering: Quasi Precedence:
a > [g, f]
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
H3(g1(x), y, z) -> H3(x, y, z)
The TRS R consists of the following rules:
f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z
The set Q consists of the following terms:
f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)
We have to consider all minimal (P,Q,R)-chains.