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(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
IMPLIES(not(x), not(y)) → IMPLIES(y, and(x, y))
IMPLIES(false, y) → NOT(false)
IMPLIES(not(x), not(y)) → AND(x, y)
IMPLIES(x, false) → NOT(x)
The TRS R consists of the following rules:
and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
IMPLIES(not(x), not(y)) → IMPLIES(y, and(x, y))
IMPLIES(false, y) → NOT(false)
IMPLIES(not(x), not(y)) → AND(x, y)
IMPLIES(x, false) → NOT(x)
The TRS R consists of the following rules:
and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
IMPLIES(not(x), not(y)) → IMPLIES(y, and(x, y))
IMPLIES(false, y) → NOT(false)
IMPLIES(not(x), not(y)) → AND(x, y)
IMPLIES(x, false) → NOT(x)
The TRS R consists of the following rules:
and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
IMPLIES(not(x), not(y)) → IMPLIES(y, and(x, y))
The TRS R consists of the following rules:
and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.