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