Logic question - just to clear up meaning of 'implies'

Question

Okay so;

I have the following statement

$p \lor \neg q \Rightarrow q \lor \neg q$

I simplify the right hand side using the complement law to get

$p \lor \neg q \Rightarrow T $

I think that $\neg q \lor p$ is the same as saying $q \Rightarrow p$ ,

so does q imply p imply true?

Will this statement always be true? I think it does; and my thinking behind it is that since q implies p, then it will either output true or false? And since the other side of the implication is always true then either false of true implies true, correct?

"I think that ¬q∨p is the same as saying q⇒p ," Yes, see Material implication rule — Mauro ALLEGRANZA, Apr 11 '22 at 10:19
"Will this statement always be true?" No, of course $q \to p$ is not always TRUE: check with truth table. Thins are different for $(q \to p) \to \top$; this is a conditional with TRUE consequent and in this case it is always TRUE. — Mauro ALLEGRANZA, Apr 11 '22 at 10:20
To the OP (i.e. original poster): I agree with your analysis. That is, if you have a statement $S$ that is always true, then any statement of the form $(R \implies S)$ can not be false (since $S$ is assumed always true). — user2661923, Apr 11 '22 at 10:21
See Logical implication as well as the related concept of Logical consequence. — Mauro ALLEGRANZA, Apr 11 '22 at 10:27

score 1 · Answer 1 · answered Apr 11 '22 at 10:19

Yes, both

$$(q \Rightarrow p) \Rightarrow T$$ and $$ (q \Rightarrow p) \Leftrightarrow (\neg q \lor p) $$ are true.

Intuitively, the value $T$ holds for any premise (since it always holds), as you said. You can check the truth table of these formulas, if in doubt.

Hope this helps

score 0 · Answer 2 · answered Apr 11 '22 at 15:45

It's kind of the dual of the principle of explosion (or vacuous truth) which states that all things follow from a falsehood:

$~~~~~~A \implies (\neg A \implies B)$

Here you have, a truth follows from all things:

$~~~~~~A\implies (B \implies A)$

Both are tautologies and are legitimate methods of proof.

ryang · Answer 3 · 2022-04-11T19:02:51.493

so does q imply p imply true?

Note that this phrasing is potentially ambiguous, because the conditional $(\to)$ is not associative, that is, $$(A → B) → C \:\not\equiv\: A → (B → C).$$ \begin{array}{ccc|c@{}c@{}c@{}ccc@{}ccc@{}ccc@{}ccc@{}ccc@{}c@{}c@{}c} a&b&c&(&(&(&a&\rightarrow&b&)&\rightarrow&c&)&\leftrightarrow&(&a&\rightarrow&(&b&\rightarrow&c&)&)&)\\\hline 1&1&1&&&&1&1&1&&1&1&&\mathbf{1}&&1&1&&1&1&1&&&\\ 1&1&0&&&&1&1&1&&0&0&&\mathbf{1}&&1&0&&1&0&0&&&\\ 1&0&1&&&&1&0&0&&1&1&&\mathbf{1}&&1&1&&0&1&1&&&\\ 1&0&0&&&&1&0&0&&1&0&&\mathbf{1}&&1&1&&0&1&0&&&\\ 0&1&1&&&&0&1&1&&1&1&&\mathbf{1}&&0&1&&1&1&1&&&\\ 0&1&0&&&&0&1&1&&0&0&&\mathbf{0}&&0&1&&1&0&0&&&\\ 0&0&1&&&&0&1&0&&1&1&&\mathbf{1}&&0&1&&0&1&1&&&\\ 0&0&0&&&&0&1&0&&0&0&&\mathbf{0}&&0&1&&0&1&0&&& \end{array}
Notice that a conditional $X\to Y$ is true whenever its consequent $Y$ is true: \begin{array}{cc|c@{}ccc@{}c} X&Y&(&X&\rightarrow&Y&)\\\hline 1&1&&1&\mathbf{1}&1&\\ 1&0&&1&\mathbf{0}&0&\\ 0&1&&0&\mathbf{1}&1&\\ 0&0&&0&\mathbf{1}&0& \end{array} Therefore, since the given proposition $$p \lor \neg q \;\to\; q \lor \neg q$$ has a tautological (always-true) consequent, it is immediately also a tautology, that is, always true. \begin{array}{cc|c@{}c@{}ccc@{}cc@{}c@{}ccc@{}ccc@{}cc@{}c@{}c@{}c} p&q&(&(&p&\lor&(&\lnot&q&)&)&\rightarrow&(&q&\lor&(&\lnot&q&)&)&)\\\hline 1&1&&&1&1&&0&1&&&\mathbf{1}&&1&1&&0&1&&&\\ 1&0&&&1&1&&1&0&&&\mathbf{1}&&0&1&&1&0&&&\\ 0&1&&&0&0&&0&1&&&\mathbf{1}&&1&1&&0&1&&&\\ 0&0&&&0&1&&1&0&&&\mathbf{1}&&0&1&&1&0&&& \end{array}

Logic question - just to clear up meaning of 'implies'

3 Answers3