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.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
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.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
+12(-2(x, y), z) -> +12(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:
POL(+12(x1, x2)) = 2·x1
POL(-2(x1, x2)) = 1 + 2·x1
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ 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.