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This question already has an answer here:

How do we explain the logical equivalence $$(P\rightarrow Q ) \equiv [(\neg P)\; \vee \; Q]$$ and if possible could you please give an example illustrating this equivalence. Thanks alot !!

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marked as duplicate by Mark Bennet, Hans Engler, user1551, colormegone, T. Bongers Jan 1 '16 at 21:37

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    $\begingroup$ Have you tried making a truth table? $\endgroup$ – Nick Sep 5 '14 at 18:58
  • $\begingroup$ Think about what it takes to make $P\rightarrow Q$ false. Then take the "not" of that to make $P\rightarrow Q$ true. $\endgroup$ – bob.sacamento Sep 5 '14 at 19:09
  • $\begingroup$ There is no such equivalence in three-valued Lukasiweicz logic. $\endgroup$ – Doug Spoonwood Sep 9 '14 at 5:16
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By definition, the implication $P\rightarrow Q$ is true whenever $P$ is false, or whenever $Q$ is true: $$\lnot P\lor Q$$

Alternatively, you can compare the truth-table for $P\rightarrow Q$ side-by-side with the truth-table for $\lnot P \lor Q$ and see that under the value in their rightmost columns, we have agreement for each and every truth-value assignment. Hence, they are logically equivalent.

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  • $\begingroup$ I'm reading the implication $(P\implies Q)$ as a mathematical writing of a statement meaning that $Q$ is true whenever $P$ is true. I'm not aware that this has a definition as you say and i can't relate this to the statement $(\bar P \;or \; Q)$ $\endgroup$ – palio Sep 5 '14 at 19:09
  • $\begingroup$ @palio Let's say I say, "If the Riemann Hypothesis is true, then the Goldbach Conjecture is true." What does this mean? It means that either the Riemann Hypothesis is false (and thus we know nothing about whether the Goldbach Conjecture is true or false), or the RH and GC are both true. In other words, either RH is false, or GC is true, or both. $\endgroup$ – Akiva Weinberger Sep 4 '15 at 23:12
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Here's an illustration:

Let $P$ be "You get a good grade on exam" and $Q$ be "I'll buy you an ice-cream". So $P $$\to$$ Q$ reads "If you get a good grade on exam, then I'll buy you an ice-cream". But that is the same as to say "Either you don't get a good grade on exam, or I'll buy you an ice-cream", which is exactly $\neg P \vee Q$. I think you can make your own illustrations similar to this argument, they all make sense in real-life language.

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  • $\begingroup$ The problem is that this $"or"$ is not exclusive, which means that i can be in the situation of not having a good grade and get an ice cream !! $\endgroup$ – palio Sep 5 '14 at 20:01
  • $\begingroup$ @palio That still corresponds to the sentence "If you get a good grade on exam, then I'll buy you an ice-cream." What if I told you, "If you get a good grade I'll buy you ice-cream, but if you don't get a good grade I could still buy you ice-cream"? $\endgroup$ – Akiva Weinberger Sep 4 '15 at 23:14
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Instead of using a truth table or a series of logical equivalences, I'd like to try a more intuitive method for you.

Try drawing a Euler diagram of $P\to Q$, as in, a small circle $P$ encompassed by a lager circle $Q$.

Now you can see, no matter where you point on this diagram, it is true you will point at $\lnot P$ or you will point within $Q$, or both.

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By definition $(P\rightarrow Q)$ is true whenever $P$ is false, or whenever $Q$ is true. Therefore : $$(P\rightarrow Q) \equiv(( \lnot P)\lor Q)$$

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You can always use truth tables of both logical statements because it is only four combinations of truth - > (T,T) (T,F) (F,T) and (F F).

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