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:
ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
The set Q consists of the following terms:
ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))
Q DP problem:
The TRS P consists of the following rules:
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(x, y)
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(x, y), 0)
The TRS R consists of the following rules:
ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
The set Q consists of the following terms:
ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(x, y)
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(x, y), 0)
The TRS R consists of the following rules:
ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
The set Q consists of the following terms:
ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
AP2(ap2(ap2(g, x), y), ap2(s, z)) -> AP2(x, y)
The TRS R consists of the following rules:
ap2(f, x) -> x
ap2(ap2(ap2(g, x), y), ap2(s, z)) -> ap2(ap2(ap2(g, x), y), ap2(ap2(x, y), 0))
The set Q consists of the following terms:
ap2(f, x0)
ap2(ap2(ap2(g, x0), x1), ap2(s, x2))
We have to consider all minimal (P,Q,R)-chains.