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