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(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, 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:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
The set Q consists of the following terms:
*2(x0, +2(x1, x2))
Q DP problem:
The TRS P consists of the following rules:
*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
The TRS R consists of the following rules:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
The set Q consists of the following terms:
*2(x0, +2(x1, x2))
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:
*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
The TRS R consists of the following rules:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
The set Q consists of the following terms:
*2(x0, +2(x1, x2))
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(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
Used argument filtering: *12(x1, x2) = x2
+2(x1, x2) = +2(x1, x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
The set Q consists of the following terms:
*2(x0, +2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.