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:
or2(x, x) -> x
and2(x, x) -> x
not1(not1(x)) -> x
not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
or2(x, x) -> x
and2(x, x) -> x
not1(not1(x)) -> x
not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> AND2(not1(x), not1(y))
NOT1(and2(x, y)) -> OR2(not1(x), not1(y))
The TRS R consists of the following rules:
or2(x, x) -> x
and2(x, x) -> x
not1(not1(x)) -> x
not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> AND2(not1(x), not1(y))
NOT1(and2(x, y)) -> OR2(not1(x), not1(y))
The TRS R consists of the following rules:
or2(x, x) -> x
and2(x, x) -> x
not1(not1(x)) -> x
not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
The TRS R consists of the following rules:
or2(x, x) -> x
and2(x, x) -> x
not1(not1(x)) -> x
not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
Used argument filtering: NOT1(x1) = x1
and2(x1, x2) = and2(x1, x2)
or2(x1, x2) = or2(x1, x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
or2(x, x) -> x
and2(x, x) -> x
not1(not1(x)) -> x
not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
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.