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