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 | [0,1,0,0] |
![](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 | [0,0] |
![](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