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) /\ (p -> r)) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [0,0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(F || ~p || q) /\ (p -> r) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [0,0,0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(F || ~p || q) /\ (p -> r) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [0,0,0,0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(~(F || p) || q) /\ (p -> r) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [0,0,0,1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || F || q) /\ (p -> r) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [0,0,1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (F || (p -> r)) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [0,0,1,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ ((F || p) -> r) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [0,0,1,1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (p -> (F || r)) => p -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (p -> r) => F || (p -> (q /\ r))
ready: no
Rule | introfalseleft |
Location | [1,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (p -> r) => (F || p) -> (q /\ r)
ready: no
Rule | introfalseleft |
Location | [1,1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (p -> r) => p -> (F || (q /\ r))
ready: no
Rule | introfalseleft |
Location | [1,1,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (p -> r) => p -> ((F || q) /\ r)
ready: no
Rule | introfalseleft |
Location | [1,1,1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (p -> r) => p -> (q /\ (F || r))
ready: no
Rule | logic.propositional.andoveror |
Location | [0,0] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p /\ (p -> r)) || (q /\ (p -> r)) => p -> (q /\ r)
ready: no
Rule | logic.propositional.defimpl |
Location | [0,0,1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (~p || r) => p -> (q /\ r)
ready: no
Rule | logic.propositional.defimpl |
Location | [1] |
![](http://ideas.cs.uu.nl/images/external.png)
(~p || q) /\ (p -> r) => ~p || (q /\ r)
ready: no