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