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:
sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(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:
SUM1(s1(x)) -> SUM1(x)
SUM1(s1(x)) -> SQR1(s1(x))
The TRS R consists of the following rules:
sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(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:
SUM1(s1(x)) -> SUM1(x)
SUM1(s1(x)) -> SQR1(s1(x))
The TRS R consists of the following rules:
sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(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:
SUM1(s1(x)) -> SUM1(x)
The TRS R consists of the following rules:
sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(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 strictly oriented and are deleted.
SUM1(s1(x)) -> SUM1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM1(x1) = SUM1(x1)
s1(x1) = s1(x1)
Lexicographic Path Order [19].
Precedence:
s1 > SUM1
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:
sum1(0) -> 0
sum1(s1(x)) -> +2(sqr1(s1(x)), sum1(x))
sqr1(x) -> *2(x, x)
sum1(s1(x)) -> +2(*2(s1(x), s1(x)), sum1(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.