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