Logic question - just to clear up meaning of 'implies' 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?
 A: 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
A: 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.
A: *

*

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}
