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:
*(x, *(y, z)) → *(*(x, y), z)
*(x, x) → x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
*(x, *(y, z)) → *(*(x, y), z)
*(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:
*1(x, *(y, z)) → *1(x, y)
*1(x, *(y, z)) → *1(*(x, y), z)
The TRS R consists of the following rules:
*(x, *(y, z)) → *(*(x, y), z)
*(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:
*1(x, *(y, z)) → *1(x, y)
*1(x, *(y, z)) → *1(*(x, y), z)
The TRS R consists of the following rules:
*(x, *(y, z)) → *(*(x, y), z)
*(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.
*1(x, *(y, z)) → *1(x, y)
*1(x, *(y, z)) → *1(*(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2) = *1(x1, x2)
*(x1, x2) = *(x1, x2)
Lexicographic path order with status [19].
Precedence:
*^12 > *2
Status:
*^12: [2,1]
*2: [2,1]
The following usable rules [14] were oriented:
*(x, *(y, z)) → *(*(x, y), z)
*(x, x) → x
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
*(x, *(y, z)) → *(*(x, y), z)
*(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.