Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(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, s(y)) → P(s(y))
-1(x, s(y)) → -1(x, p(s(y)))
The TRS R consists of the following rules:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
-1(x, s(y)) → P(s(y))
-1(x, s(y)) → -1(x, p(s(y)))
The TRS R consists of the following rules:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
-1(x, s(y)) → P(s(y))
-1(x, s(y)) → -1(x, p(s(y)))
The TRS R consists of the following rules:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → 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
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-1(x, s(y)) → -1(x, p(s(y)))
The TRS R consists of the following rules:
-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.