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:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
H2(z, e1(x)) -> H2(c1(z), d2(z, x))
D2(c1(z), g2(g2(x, y), 0)) -> G2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
D2(z, g2(x, y)) -> D2(z, y)
H2(z, e1(x)) -> D2(z, x)
G2(e1(x), e1(y)) -> G2(x, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
D2(z, g2(x, y)) -> G2(e1(x), d2(z, y))
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
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:
H2(z, e1(x)) -> H2(c1(z), d2(z, x))
D2(c1(z), g2(g2(x, y), 0)) -> G2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
D2(z, g2(x, y)) -> D2(z, y)
H2(z, e1(x)) -> D2(z, x)
G2(e1(x), e1(y)) -> G2(x, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
D2(z, g2(x, y)) -> G2(e1(x), d2(z, y))
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G2(e1(x), e1(y)) -> G2(x, y)
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
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.
G2(e1(x), e1(y)) -> G2(x, y)
Used argument filtering: G2(x1, x2) = x2
e1(x1) = e1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
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
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
D2(z, g2(x, y)) -> D2(z, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
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.
D2(z, g2(x, y)) -> D2(z, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))
Used argument filtering: D2(x1, x2) = x2
g2(x1, x2) = g2(x1, x2)
0 = 0
e1(x1) = e
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
H2(z, e1(x)) -> H2(c1(z), d2(z, x))
The TRS R consists of the following rules:
h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.