Definition talk:Therefore

I honestly have yet to figure out how this is different from the conditional. Is there a simple explanation, or do I just need to take a course in mathematical logic? --Cynic (talk) 03:25, 27 June 2009 (UTC)

Not sure there really is one ultimately. H. Jerome Keisler and Joel Robbin's "Mathematical Logic and Computability" doesn't even mention the "therefore" concept, but their approach is unusual.

In a slightly different context, but with the same message (and I've translated the symbols to match what we have on this site), M. Ben-Ari's "Mathematical Logic for Computer Science" says: "Isn't $$\dashv \vdash$$ a boolean operator? The answer is no. It is simply a shorthand for the phrase "is logically equivalent to", unlike $$\iff$$ which is a boolean operator in the logic that we are describing."

I'll get back to you on this, I have a domestic duty to perform ... --Matt Westwood 10:07, 27 June 2009 (UTC)

Back again. Another thing: $$\neg \left({p \implies q}\right)$$ has a specifically defined behaviour which is the negation of $$\implies$$, which is true only when $$p$$ is true and $$q$$ is false, whereas $$p \not \vdash q$$ just means "$$p$$ does not imply $$q$$, that is, it does not necessarily follow that when $$p$$ is true, then $$q$$ must be false.

Easier to follow what I mean when I discuss the difference between $$p \iff q$$ and $$p \dashv \vdash q$$.

Again, $$\neg \left({p \iff q}\right)$$ is the exclusive or operator, whereas the negation of $$p \dashv \vdash q$$ does not necessarily mean that $$p$$ and $$q$$ have opposite truth values, it just mean it doesn't automatically have to be the case that they are the same. --Matt Westwood 10:24, 27 June 2009 (UTC)