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I see the symbol used, and I've never seen it logically defined. In words, It's defined as a symbol meaning 'therefore'. Because of a lack of definition, I have no idea why this is false:

A
B
∴C

For me, this statemnet could mean one of two things.

  1. 'If A and B thus C': (A^B) -> C i.e same as conditional

  2. That ∴ is the same as the conditional, only without a truth value for when the premises is false (conclusions cannot draw upon a false premise) 'A thus B', perhaps? I've read that a valid argument is only so if it is (logically/symbolically/algebraically) impossible that the conclusion be false, given that premise is true.

I am not certain on that, and any correction would be appreciated. However, regardless, I still can't find a strict, logical definition or equivalent of ∴, and it's not that I haven't looked, the ∴ only ever seems to be talked about, never strictly defined.

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    $\begingroup$ Basically, 1. It just means "therefore" :) $\endgroup$ – BrianO May 12 '16 at 1:52
  • $\begingroup$ Oh ok Sweet. :) $\endgroup$ – Jim Jam May 12 '16 at 1:54
  • $\begingroup$ In older books you may see the upside-down version of that symbol, meaning "since" or "because". $\endgroup$ – DanielWainfleet May 12 '16 at 2:02
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The symbol means "therefore". In common usage, "therefore" is a stronger statement than "if/then". When we say "p therefore q", we mean both "if p then q" and "p is true". Thus we assure that q is true, which the "if/then" statement alone does not. Mathematically:

$p$ therefore $q \equiv ((p\rightarrow q) and p)$

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  • $\begingroup$ The proposition $(p\to q)\wedge p$ is logically equivalent to $p\wedge q$. This solution overlooks the semantic meaning of "therefore". $\endgroup$ – vadim123 May 12 '16 at 1:59
  • $\begingroup$ Ahh, so it's not just a confusing term, with no symbolic meaning. Though I do see why the therefore and and symbols, even though logically the same, would differ, as the former represents a step in an premise, where the latter represents the entire premise and concluding point :) $\endgroup$ – Jim Jam May 12 '16 at 2:13
  • $\begingroup$ Or you could say that "$p$ and $q$ is true" is a corrollary if you prove both "if $p$ then $q$" and "$p$ is true". $\endgroup$ – Oscar Lanzi May 12 '16 at 2:14
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It is a semantic statement rather than a syntactic one. Syntax is the level of propositional calculus in which $A,B, A\wedge B$ live. Semantics is at a higher level, where we assign truth values to propositions based on interpreting them in a larger universe.

Your (1), $(A\wedge B) \to C$, is a proposition. It may be true or false. However $(A\wedge B) \therefore C$ cannot be false. It can be a valid proof, or invalid, which is again at a a semantic level rather than a syntactic one.

For example, if $P$ holds then $\neg(\neg P)$ must hold. We have $P \therefore \neg(\neg P)$. This is a different meaning than $P\to \neg(\neg P)$, which is a proposition that happens to be always true (a tautology). Asserting that $P\to \neg(\neg P)$ is logically equivalent to a tautology, is equivalent to $P \therefore \neg(\neg P)$.

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  • $\begingroup$ But there must be a way for us to know if it's valid or invalid, the whole point of logic, surely, is to eliminate interpretation. $\endgroup$ – Jim Jam May 12 '16 at 1:58
  • $\begingroup$ To test if a proof is valid we must exit propositional calculus. For example, using a truth table. In practice, we prove certain theorems semantically using truth tables (like modus ponens), and then use them to prove other things. $\endgroup$ – vadim123 May 12 '16 at 2:00
  • $\begingroup$ @user108262: You remember math.stackexchange.com/a/1684204/21820? Just look up the rules for Fitch-style natural deduction.. there are only a small number, one introduction rule and one elimination rule for each connective, and then add the rules I gave you for quantifiers and you're done. Any argument that cannot be proven via those rules is invalid. Simple as that. $\endgroup$ – user21820 May 12 '16 at 6:44
  • $\begingroup$ @Though I must say that this fact that those rules are sufficient is by no means trivial. It's known as the completeness theorem for first-order logic, and requires a strong enough meta-system to prove this fact. $\endgroup$ – user21820 May 12 '16 at 6:49

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