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:
g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G2(x, s1(y)) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)
The TRS R consists of the following rules:
g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))
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:
G2(x, s1(y)) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)
The TRS R consists of the following rules:
g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G2(f2(x, y), 0) -> G2(y, 0)
G2(s1(x), y) -> G2(f2(x, y), 0)
G2(f2(x, y), 0) -> G2(x, 0)
The TRS R consists of the following rules:
g2(0, f2(x, x)) -> x
g2(x, s1(y)) -> g2(f2(x, y), 0)
g2(s1(x), y) -> g2(f2(x, y), 0)
g2(f2(x, y), 0) -> f2(g2(x, 0), g2(y, 0))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.