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:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
Q is empty.
We have applied [15,7] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
The signature Sigma is {h}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
The set Q consists of the following terms:
h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
H(x, c(y, z)) → H(c(s(y), x), z)
The TRS R consists of the following rules:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
The set Q consists of the following terms:
h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
H(x, c(y, z)) → H(c(s(y), x), z)
The TRS R consists of the following rules:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
The set Q consists of the following terms:
h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)
We have to consider all minimal (P,Q,R)-chains.