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