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