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:
not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.