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Jun 5, 2016 at 15:50 comment added Ovi Thanks a lot, your last comment (especially the last sentence) helped A LOT!
Jun 5, 2016 at 15:44 vote accept Ovi
Jun 5, 2016 at 8:14 history edited Mauro ALLEGRANZA CC BY-SA 3.0
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Jun 5, 2016 at 7:36 comment added Mauro ALLEGRANZA @Ovi - in math usually we are not interested into conditional whatever like "if $0=1$, then the moon is round", but to theorems proved from axioms. Thus, the conditional we are using are like: "if axiom $A$ holds, then theorem $T$ holds also". The fact that the conditional is true also when $A$ is false is a fact than we accept but of course is of no relevance for the mathematical theory we are workin whit.
Jun 4, 2016 at 20:44 comment added Ovi ...Does it mean that it's not valid to put $p$ and $q$ in a $p \implies q$ relationship? Does it mean it's to put in in $p \implies q$ as long as $q$ as long as $q$ is true?
Jun 4, 2016 at 20:43 comment added Ovi So my understanding is that when we are given a theorem in math as an $p \implies q$ statement, we are being told "This $p \implies w$ statement is always true, meaning that we can possibly find every combination of $p$ and $q$ except for $p=F$ and $q=T$"? I guess what I'm not really sure I get why my example here (math.stackexchange.com/questions/1793713/…) does not work (would really appreciate it if you could take a look). As an answer to that question, people said $p$ and $q$ were not independent. But what does that mean?
Jun 4, 2016 at 19:08 history edited Mauro ALLEGRANZA CC BY-SA 3.0
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Jun 4, 2016 at 19:02 history answered Mauro ALLEGRANZA CC BY-SA 3.0