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))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, 1) -> x
*2(1, y) -> y
Q is empty.
↳ 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))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, 1) -> x
*2(1, y) -> y
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
*12(+2(x, y), z) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
*12(+2(x, y), z) -> *12(y, z)
The TRS R consists of the following rules:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, 1) -> x
*2(1, y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
*12(+2(x, y), z) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
*12(+2(x, y), z) -> *12(y, z)
The TRS R consists of the following rules:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, 1) -> x
*2(1, y) -> 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.
*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: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
*12(+2(x, y), z) -> *12(x, z)
*12(+2(x, y), z) -> *12(y, z)
The TRS R consists of the following rules:
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, 1) -> x
*2(1, y) -> 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.
*12(+2(x, y), z) -> *12(x, z)
*12(+2(x, y), z) -> *12(y, z)
Used argument filtering: *12(x1, x2) = x1
+2(x1, x2) = +2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ 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))
*2(+2(x, y), z) -> +2(*2(x, z), *2(y, z))
*2(x, 1) -> x
*2(1, y) -> 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.