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I am a bit confused about this truth table that I have copied down during my lessons. The question is

  • Use a truth table to determine whether the statement (P→Q)∧(Q→P) is logically equivalent to (P∨Q)∧(¬P∨¬Q)

This is the truth table that I have written down: enter image description here

What I am confused now is if the fourth column is it really needed (⇔) for this question?

While compiling my notes, there is another similar question where it asks if (P→Q) is logically equivalent to ¬P∨Q and that particular column (⇔) is not shown in the truth table, I assume that is because the result of (P→Q) and ¬P∨Q is the same, hence enough said?

And so, is that column needed? Otherwise, under what sort of scenario or how are the questions being phrased that would requires me to use ⇔?

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  • $\begingroup$ I think the idea is that you put '1' in the $\iff$ column whenever the relevant entries match; if it's all 1's, then the two are indeed equivalent. But I've never seen it before, it's not necessary by any means. $\endgroup$
    – pjs36
    Commented Mar 30, 2016 at 5:50
  • $\begingroup$ do you mean to say that ⇔ only applies for cases where the outputs are only 1's? I remember that if all outputs are 1's, they are called tautology. But what about if they are all 0's? $\endgroup$
    – dissidia
    Commented Mar 30, 2016 at 6:36
  • $\begingroup$ I mean this: If you look at the two columns we're comparing, $(P \to Q) \land (Q \to P)$ and $(P \lor Q) \land (\neg P \lor \neg Q)$, we put '1' in the $\iff$ column whenever their values match, and '0' whenever they don't match. In the example you have here, they're always opposites, so the $\iff$ column is always '0'. At least, that's what I assume is happening. $\endgroup$
    – pjs36
    Commented Mar 30, 2016 at 6:43
  • $\begingroup$ If they are all 0's, it's a contradiction. So no matter what the values of your two expressions are, it will ALWAYS be false. $\endgroup$
    – Inazuma
    Commented Mar 30, 2016 at 6:49

1 Answer 1

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Two expressions are logically equivalent if and only if their truth values are equivalent. The first two expressions you listed (P→Q)∧(Q→P), (P∨Q)∧(¬P∨¬Q) are therefore not equivalent.

The second pair of expressions, (P→Q) and ¬P∨Q however, are equal (as you correctly determined). In fact, one definition of (P→Q) is ¬P∨Q.

The double arrow means if and only if, so that if you have a statement p ⇔ q, then you need to show p → q AND q → p (which is coincidentally logically equivalent to the third column you've written). It is unnecessary for determining whether two expressions are logically equivalent though, nor can I see the relation in the example.

Reading a quick excerpt for the first book I could find under the search "double arrow logical equivalence":

"Another way to determine whether two propositions are logical equivalent is to join two propositions together using the double arrow (⇔) and then determine whether the resulting biconditional forms a tautology... thus, if a biconditional is a tautology, then the left and right sides of the biconditional are logically equivalent to each other".

Source: Symbolic Logic: Syntax, Semantics, and Proof By David W. Agler, Pg. 82, Chapter 3

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  • $\begingroup$ do correct me if I am wrong. Using the question I have stated - if the outputs of (P→Q)∧(Q→P) and (P∨Q)∧(¬P∨¬Q), there is not a need for the double arrows (⇔). Whereas unless if the outputs are the same (the second example I have used), I can perhaps then put in the double arrows into a statement such as Therefore (P→Q) ⇔ ¬P∨Q, the column is just unnecessary in both cases then $\endgroup$
    – dissidia
    Commented Mar 30, 2016 at 6:35
  • $\begingroup$ I have done some further research and provided a source of the method your teacher may have been adopting. I hope this is satisfactory. $\endgroup$
    – Inazuma
    Commented Mar 30, 2016 at 6:54
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    $\begingroup$ Thanks! I think I just drop my teacher an email or look for more questions as to when such scenarios will it be used $\endgroup$
    – dissidia
    Commented Mar 30, 2016 at 7:22

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