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:
and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
IMPLIES2(not1(x), not1(y)) -> IMPLIES2(y, and2(x, y))
IMPLIES2(false, y) -> NOT1(false)
IMPLIES2(not1(x), not1(y)) -> AND2(x, y)
IMPLIES2(x, false) -> NOT1(x)
The TRS R consists of the following rules:
and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, 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:
IMPLIES2(not1(x), not1(y)) -> IMPLIES2(y, and2(x, y))
IMPLIES2(false, y) -> NOT1(false)
IMPLIES2(not1(x), not1(y)) -> AND2(x, y)
IMPLIES2(x, false) -> NOT1(x)
The TRS R consists of the following rules:
and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IMPLIES2(not1(x), not1(y)) -> IMPLIES2(y, and2(x, y))
The TRS R consists of the following rules:
and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.