# Logic - What does ∴ mean in a truth table?

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.

• Basically, 1. It just means "therefore" :) – BrianO May 12 '16 at 1:52
• Oh ok Sweet. :) – Jim Jam May 12 '16 at 1:54
• In older books you may see the upside-down version of that symbol, meaning "since" or "because". – DanielWainfleet May 12 '16 at 2:02

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

• The proposition $(p\to q)\wedge p$ is logically equivalent to $p\wedge q$. This solution overlooks the semantic meaning of "therefore". – vadim123 May 12 '16 at 1:59
• 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 :) – Jim Jam May 12 '16 at 2:13
• 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". – Oscar Lanzi May 12 '16 at 2:14

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 $P \therefore \neg(\neg P)$.

• 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. – Jim Jam May 12 '16 at 1:58
• 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. – vadim123 May 12 '16 at 2:00
• @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. – user21820 May 12 '16 at 6:44
• @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. – user21820 May 12 '16 at 6:49