Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
:12(z, +2(x, f1(y))) -> :12(g2(z, y), +2(x, a))
:12(:2(x, y), z) -> :12(y, z)
:12(:2(x, y), z) -> :12(x, :2(y, z))
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)
The TRS R consists of the following rules:
:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))
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:
:12(z, +2(x, f1(y))) -> :12(g2(z, y), +2(x, a))
:12(:2(x, y), z) -> :12(y, z)
:12(:2(x, y), z) -> :12(x, :2(y, z))
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)
The TRS R consists of the following rules:
:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
:12(:2(x, y), z) -> :12(x, :2(y, z))
:12(:2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)
The TRS R consists of the following rules:
:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))
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(:2(x, y), z) -> :12(x, :2(y, z))
:12(:2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)
Used argument filtering: :12(x1, x2) = x1
:2(x1, x2) = :2(x1, x2)
+2(x1, x2) = +2(x1, x2)
f1(x1) = f
g2(x1, x2) = g
a = a
Used ordering: Precedence:
:2 > +2 > g
f > a > g
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))
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.