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(+2(x, z), y)
-2(+2(x, y), y) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(x, y), y) -> x
Q is empty.
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(+2(x, z), y)
The TRS R consists of the following rules:
+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(x, y), y) -> x
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(x, z)
+12(-2(x, y), z) -> -12(+2(x, z), y)
The TRS R consists of the following rules:
+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(x, y), y) -> x
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, z)
The TRS R consists of the following rules:
+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(x, y), y) -> x
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)
Used argument filtering: +12(x1, x2) = x1
-2(x1, x2) = -1(x1)
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), z) -> -2(+2(x, z), y)
-2(+2(x, y), y) -> x
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.