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:
H1(C(S(x), C(S(0), y)), z) → H1(y, C(S(0), C(x, z)))
H1(x, C(y, z)) → H1(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:
H1(C(S(x), C(S(0), y)), z) → H1(y, C(S(0), C(x, z)))
H1(x, C(y, z)) → H1(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.