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Basic logic question

I'm wondering if using a truth table to prove

p⇒q≡¬p∨qp\Rightarrow q \equiv \neg p \vee q

would be circular logic?


My guess is no but when I think of the truth table for p⇒qp\Rightarrow q, I think "either q is true or p is false", so I can fill in the table using this. But that's what I'm trying to prove in the first place.

So maybe my thought process is circular but my proof isn't?
Original post by 0-)
I'm wondering if using a truth table to prove

p⇒q≡¬p∨qp\Rightarrow q \equiv \neg p \vee q

would be circular logic?


My guess is no but when I think of the truth table for p⇒qp\Rightarrow q, I think "either q is true or p is false", so I can fill in the table using this. But that's what I'm trying to prove in the first place.

So maybe my thought process is circular but my proof isn't?

Depends how you've defined the "or" symbol. What have you defined?
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Original post by Smaug123
Depends how you've defined the "or" symbol. What have you defined?

I'm new to logic so my only definition would be with a truth table,

p, q, (p V q)
T, T, T
T, F, T
F, T, T
F, F, F

Can you explain how the definition would change whether the proof is circular?
Original post by 0-)
I'm new to logic so my only definition would be with a truth table,

p, q, (p V q)
T, T, T
T, F, T
F, T, T
F, F, F

Can you explain how the definition would change whether the proof is circular?

In my course, we defined p∨qp \vee q to be (¬p)⇒q(\lnot p) \Rightarrow q. It's conceivable that you could have defined "implies" in terms of "or" or something.

If you defined "or" as a truth table, then your reasoning is fine. If we let vv be a valuation on a language in which p,qp, q are propositions, then v(p⇒q)v(p \Rightarrow q) is 1 unless v(p)=1,v(q)=0v(p) = 1, v(q) = 0, when it's 0: that is, 1 unless a true thing is implying a false thing. v(p∨q)v(p \vee q) is defined in the same way as v((¬p)⇒q)v((\lnot p) \Rightarrow q), according to your definition, so that's a proof that they are indeed the same.

Feel free to ignore this next paragraph unless you've heard of the Completeness Theorem.

It appears that you've defined "or" in terms of what valuations do to the statement, by the way: that's implicitly relying on the Completeness Theorem, which states that a proposition of first-order logic is semantically true iff it is syntactically true. (That is, "I can examine how the statement behaves as a truth table, and that tells me about precisely whether the statement can be proved by symbol manipulation".) Depends how you've defined "implies", as well.

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