Termination w.r.t. Q of the following Term Rewriting System could be proven:
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
↳ QDPOrderProof
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.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
AND(not(not(x)), y, not(z)) → AND(y, band(x, z), x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
AND(x1, x2, x3) = AND(x1, x2)
not(x1) = not
band(x1, x2) = band
Recursive path order with status [2].
Precedence:
not > AND2
not > band
Status:
not: multiset
band: multiset
AND2: multiset
The following usable rules [14] were oriented:
none
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.