Exercise logic.propositional.consequence
Description
Prove that formula is a logical consequence of a set of formulas
All applications
![](http://ideas.cs.uu.nl/images/external.png)
(p <-> r) /\ (q <-> s) => (~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
F || (p <-> r), q <-> s => (~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(F || p) <-> r, q <-> s => (~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,0,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> (F || r), q <-> s => (~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, F || (q <-> s) => (~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,1,0] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, (F || q) <-> s => (~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [0,1,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> (F || s) => (~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => F || ((~p || q) <-> (r -> s))
ready: no
Rule | introfalseleft |
Location | [1,0] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (F || ~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1,0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (F || ~p || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1,0,0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (~(F || p) || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1,0,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (~p || F || q) <-> (r -> s)
ready: no
Rule | introfalseleft |
Location | [1,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (~p || q) <-> (F || (r -> s))
ready: no
Rule | introfalseleft |
Location | [1,1,0] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (~p || q) <-> ((F || r) -> s)
ready: no
Rule | introfalseleft |
Location | [1,1,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (~p || q) <-> (r -> (F || s))
ready: no
Rule | logic.propositional.defequiv |
Location | [0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(p /\ r) || (~p /\ ~r), q <-> s => (~p || q) <-> (r -> s)
ready: no
Rule | logic.propositional.defequiv |
Location | [0,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, (q /\ s) || (~q /\ ~s) => (~p || q) <-> (r -> s)
ready: no
Rule | logic.propositional.defequiv |
Location | [1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => ((~p || q) /\ (r -> s)) || (~(~p || q) /\ ~(r -> s))
ready: no
Rule | logic.propositional.defimpl |
Location | [1,1] |
![](http://ideas.cs.uu.nl/images/external.png)
p <-> r, q <-> s => (~p || q) <-> (~r || s)
ready: no