# Discrete Math - p imp q Truth Table

I'm trying to understand what imp1 is or how it is used. I'm unable to find further information in my textbook and this is an exercise question.

Question: Provide further motivation for defining p → q to be true when p is false For the first change, we call the resulting operator imp1.

Show that p imp1 q logically equivalent q imp1 p.

(BTW if anyone can direct me to the help section on how to use math symbols in a post I would greatly appreciate it so these questions make more sense).

Thanks all. • As a note I understand that they are logically equivalent if the truth tables are identical when showing the converse of p imp q. Again I just need some clarification on what imp is. – natur3 Jan 15 '15 at 4:07
• After a little more reading am I correct in concluding that imp is a word phrase used in place of whatever conditional operation was used? – natur3 Jan 15 '15 at 4:09
• We have two connectives in palce : the conditional (or implication) : $\rightarrow$ (written in LaTex : \rightarrow) and the bi-conditional (or bi-implication or equivalence) : $\leftrightarrow$ (written in LaTex : \leftrightarrow) . The truth-table for the first one (in case : $p \rightarrow q$) has $FALSE$ only in the row with $p - FALSE$ and $q - TRUE$. The one in your post (called "imp1") is the truth-table for $p \leftrightarrow q$; it has $FALSE$ is the rows where $p$ and $q$ have different truth values. – Mauro ALLEGRANZA Jan 15 '15 at 8:30
• For a tutorial on math formulae formatting, see here – Mauro ALLEGRANZA Jan 15 '15 at 8:35
• @MauroALLEGRANZA thanks for this breakdown it makes great sense! – natur3 Jan 16 '15 at 3:20

Since $p \; imp1 \; q$ is given by the truth table in your question, we see that \begin{array}{| l | l | l | l |} \hline p & q & p \; imp1 \; q & q \; imp1 \; p \\ \hline T & T & T & T \\ \hline T & F & F & F \\ \hline F & T & F & F \\ \hline F & F & T & T \\ \hline \end{array} so $p \; imp1 \; q$ is logically equivalent to $q \; imp1 \; p$. Notice that $p \; imp1 \; q$ is true only when $p$ and $q$ have the same truth value. This operator is usually called bi-implication and is written $\leftrightarrow$ and is often read as "if and only if".
• @natur3 the converse of $p \rightarrow q$ is $q \rightarrow p$ which are not logically equivalent. You are right however, that $p \leftrightarrow q$ is logically equivalent to $(p \rightarrow q) \wedge (q \rightarrow p)$, that is, to show bi-implication we need to show implication in both directions. I hope this helped. – mrp Jan 16 '15 at 10:17