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:
and(not(not(x)), y, not(z)) → and(y, band(x, z), x)
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
and(not(not(x)), y, not(z)) → and(y, band(x, z), x)
Q is empty.
We have applied [15,7] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
and(not(not(x)), y, not(z)) → and(y, band(x, z), x)
The signature Sigma is {and}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
and(not(not(x)), y, not(z)) → and(y, band(x, z), x)
The set Q consists of the following terms:
and(not(not(x0)), x1, not(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:
AND(not(not(x)), y, not(z)) → AND(y, band(x, z), x)
The TRS R consists of the following rules:
and(not(not(x)), y, not(z)) → and(y, band(x, z), x)
The set Q consists of the following terms:
and(not(not(x0)), x1, not(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:
AND(not(not(x)), y, not(z)) → AND(y, band(x, z), x)
The TRS R consists of the following rules:
and(not(not(x)), y, not(z)) → and(y, band(x, z), x)
The set Q consists of the following terms:
and(not(not(x0)), x1, not(x2))
We have to consider all minimal (P,Q,R)-chains.