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 <-> r) /\ (q <-> s)) => (p -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,0] |
(F || (p <-> r)) /\ (q <-> s) => (p -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,0,0] |
((F || p) <-> r) /\ (q <-> s) => (p -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,0,1] |
(p <-> (F || r)) /\ (q <-> s) => (p -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,1] |
(p <-> r) /\ (F || (q <-> s)) => (p -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,1,0] |
(p <-> r) /\ ((F || q) <-> s) => (p -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,1,1] |
(p <-> r) /\ (q <-> (F || s)) => (p -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1] |
(p <-> r) /\ (q <-> s) => F || ((p -> q) <-> (r -> s))
ready: no
Rule | introfalseleft |
Location | [1,0] |
(p <-> r) /\ (q <-> s) => (F || (p -> q)) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1,0,0] |
(p <-> r) /\ (q <-> s) => ((F || p) -> q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1,0,1] |
(p <-> r) /\ (q <-> s) => (p -> (F || q)) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1,1] |
(p <-> r) /\ (q <-> s) => (p -> q) <-> (F || (r -> s))
ready: no
Rule | introfalseleft |
Location | [1,1,0] |
(p <-> r) /\ (q <-> s) => (p -> q) <-> ((F || r) -> s)
ready: no
Rule | introfalseleft |
Location | [1,1,1] |
(p <-> r) /\ (q <-> s) => (p -> q) <-> (r -> (F || s))
ready: no
Rule | logic.propositional.defequiv |
Location | [0,0,0] |
((p /\ r) || (~p /\ ~r)) /\ (q <-> s) => (p -> q) <-> (r -> s)
ready: no
Rule | logic.propositional.defequiv |
Location | [0,0,1] |
(p <-> r) /\ ((q /\ s) || (~q /\ ~s)) => (p -> q) <-> (r -> s)
ready: no
Rule | logic.propositional.defequiv |
Location | [1] |
(p <-> r) /\ (q <-> s) => ((p -> q) /\ (r -> s)) || (~(p -> q) /\ ~(r -> s))
ready: no
Rule | logic.propositional.defimpl |
Location | [1,0] |
(p <-> r) /\ (q <-> s) => (~p || q) <-> (r -> s)
ready: no
Rule | logic.propositional.defimpl |
Location | [1,1] |
(p <-> r) /\ (q <-> s) => (p -> q) <-> (~r || s)
ready: no