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), z)
*2(x, x) -> x
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), z)
*2(x, x) -> 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(x, *2(y, z)) -> *12(x, y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)
The TRS R consists of the following rules:
*2(x, *2(y, z)) -> *2(*2(x, y), z)
*2(x, x) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
*12(x, *2(y, z)) -> *12(x, y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)
The TRS R consists of the following rules:
*2(x, *2(y, z)) -> *2(*2(x, y), z)
*2(x, x) -> 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(x, *2(y, z)) -> *12(x, y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:
POL(*2(x1, x2)) = 3 + x1 + x2
POL(*12(x1, x2)) = 2·x1 + 3·x2
The following usable rules [14] were oriented:
*2(x, *2(y, z)) -> *2(*2(x, y), z)
*2(x, x) -> x
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ 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), z)
*2(x, x) -> 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.