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(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
F2(x, g1(y)) -> I2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> J2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
I2(x, j2(y, z)) -> J2(g1(y), i2(x, z))
J2(g1(x), g1(y)) -> J2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)
F2(x, g1(y)) -> F2(h1(x), i2(x, y))
The TRS R consists of the following rules:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(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:
I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
F2(x, g1(y)) -> I2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> J2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
I2(x, j2(y, z)) -> J2(g1(y), i2(x, z))
J2(g1(x), g1(y)) -> J2(x, y)
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)
F2(x, g1(y)) -> F2(h1(x), i2(x, y))
The TRS R consists of the following rules:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(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:
J2(g1(x), g1(y)) -> J2(x, y)
The TRS R consists of the following rules:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(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.
J2(g1(x), g1(y)) -> J2(x, y)
Used argument filtering: J2(x1, x2) = x2
g1(x1) = g1(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:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(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:
I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)
The TRS R consists of the following rules:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(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.
I2(h1(x), j2(j2(y, z), 0)) -> I2(h1(x), j2(y, z))
I2(h1(x), j2(j2(y, z), 0)) -> I2(x, j2(y, z))
I2(x, j2(y, z)) -> I2(x, z)
Used argument filtering: I2(x1, x2) = x2
j2(x1, x2) = j2(x1, x2)
0 = 0
g1(x1) = g
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:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(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:
F2(x, g1(y)) -> F2(h1(x), i2(x, y))
The TRS R consists of the following rules:
f2(x, g1(y)) -> f2(h1(x), i2(x, y))
i2(x, j2(0, 0)) -> g1(0)
i2(x, j2(y, z)) -> j2(g1(y), i2(x, z))
i2(h1(x), j2(j2(y, z), 0)) -> j2(i2(h1(x), j2(y, z)), i2(x, j2(y, z)))
j2(g1(x), g1(y)) -> g1(j2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.