Exercise logic.propositional.consequence
Description
Prove that formula is a logical consequence of a set of formulas
All applications
Rule | introfalseleft |
Location | [0,0] |
F || (~p /\ (q -> r)) || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,0] |
F || (~p /\ (q -> r)) || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,0,0] |
((F || ~p) /\ (q -> r)) || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,0,0,0] |
(~(F || p) /\ (q -> r)) || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,0,1] |
(~p /\ (F || (q -> r))) || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,0,1,0] |
(~p /\ ((F || q) -> r)) || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,0,1,1] |
(~p /\ (q -> (F || r))) || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,1] |
(~p /\ (q -> r)) || F || (q /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,1,0] |
(~p /\ (q -> r)) || ((F || q) /\ (q -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,1,1] |
(~p /\ (q -> r)) || (q /\ (F || (q -> r))) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,1,1,0] |
(~p /\ (q -> r)) || (q /\ ((F || q) -> r)) => p -> r
ready: no
Rule | introfalseleft |
Location | [0,0,1,1,1] |
(~p /\ (q -> r)) || (q /\ (q -> (F || r))) => p -> r
ready: no
Rule | introfalseleft |
Location | [1] |
(~p /\ (q -> r)) || (q /\ (q -> r)) => F || (p -> r)
ready: no
Rule | introfalseleft |
Location | [1,0] |
(~p /\ (q -> r)) || (q /\ (q -> r)) => (F || p) -> r
ready: no
Rule | introfalseleft |
Location | [1,1] |
(~p /\ (q -> r)) || (q /\ (q -> r)) => p -> (F || r)
ready: no
Rule | logic.propositional.defimpl |
Location | [0,0,0,1] |
(~p /\ (~q || r)) || (q /\ (q -> r)) => p -> r
ready: no
Rule | logic.propositional.defimpl |
Location | [0,0,1,1] |
(~p /\ (q -> r)) || (q /\ (~q || r)) => p -> r
ready: no
Rule | logic.propositional.defimpl |
Location | [1] |
(~p /\ (q -> r)) || (q /\ (q -> r)) => ~p || r
ready: no
Rule | logic.propositional.oroverand |
Location | [0,0] |
((~p /\ (q -> r)) || q) /\ ((~p /\ (q -> r)) || (q -> r)) => p -> r
ready: no
Rule | logic.propositional.oroverand |
Location | [0,0] |
(~p || (q /\ (q -> r))) /\ ((q -> r) || (q /\ (q -> r))) => p -> r
ready: no