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), *2(a, y)) -> *2(+2(x, a), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(*2(x, y), *2(a, y)) -> *2(+2(x, a), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
*12(*2(x, y), z) -> *12(y, z)
+12(*2(x, y), *2(a, y)) -> *12(+2(x, a), y)
+12(*2(x, y), *2(a, y)) -> +12(x, a)
*12(*2(x, y), z) -> *12(x, *2(y, z))
The TRS R consists of the following rules:
+2(*2(x, y), *2(a, y)) -> *2(+2(x, a), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
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(*2(x, y), z) -> *12(y, z)
+12(*2(x, y), *2(a, y)) -> *12(+2(x, a), y)
+12(*2(x, y), *2(a, y)) -> +12(x, a)
*12(*2(x, y), z) -> *12(x, *2(y, z))
The TRS R consists of the following rules:
+2(*2(x, y), *2(a, y)) -> *2(+2(x, a), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
*12(*2(x, y), z) -> *12(y, z)
*12(*2(x, y), z) -> *12(x, *2(y, z))
The TRS R consists of the following rules:
+2(*2(x, y), *2(a, y)) -> *2(+2(x, a), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
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(y, z)
*12(*2(x, y), z) -> *12(x, *2(y, z))
Used argument filtering: *12(x1, x2) = x1
*2(x1, x2) = *2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ 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), *2(a, y)) -> *2(+2(x, a), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
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.