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:
-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(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, s1(y)) -> P1(s1(y))
-12(x, s1(y)) -> -12(x, p1(s1(y)))
The TRS R consists of the following rules:
-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> 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(x, s1(y)) -> P1(s1(y))
-12(x, s1(y)) -> -12(x, p1(s1(y)))
The TRS R consists of the following rules:
-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(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
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-12(x, s1(y)) -> -12(x, p1(s1(y)))
The TRS R consists of the following rules:
-2(0, y) -> 0
-2(x, 0) -> x
-2(x, s1(y)) -> if3(greater2(x, s1(y)), s1(-2(x, p1(s1(y)))), 0)
p1(0) -> 0
p1(s1(x)) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.