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It seems you have seen this question posed by many users. But still i want to clear my doubt Regarding Implication in Mathematical Logic which is $p \to q$

I have clearly understood these:

$1.$ If $p$ is True, $q$ is True, Then the Implication is True

$2.$ If $p$ is false, $q$ is False, Then the Implication is True

$3.$ If $p$ is True, $q$ is False, Then the Implication is False

But why If $p$ is False and $q$ is True, The Implication is treated as True.

One example I convinced myself with is:

If $x=2$, Then $x^2+1=5$

Where $p$ is the statement $x=2$ and $q$ is the statement $x^2+1=5$

Now if $p$ is false, Then still $q$ can be true, since $x=-2$ will help to make $q$ True.But there are infinite values of $x$ which makes $q$ false, hence majority of values of $x$ are making $q$ false. Hence the Implication is TRUE.

Now i started to think about a different example Viz:

If $x \gt 7$, Then $x \gt 5$

Now if $p$ is false There are infinite values of $x$ making $q$ True that is all numbers $ x \in (5 \:\: 7]$

Also $q$ is false for all $x \in (-\infty \:\: 5]$

Since density of numbers in second interval is more, The Implication Should be TRUE again.

Is this the Reason when $p$ is False and $q$ is True, Implication is considered TRUE?, since when $p$ is false the number of reasons for $q$ being true less than number of causes making $q$ false?

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marked as duplicate by Namaste discrete-mathematics Feb 8 '18 at 17:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The key idea here is that we are referring to the entire conditional when we determine its truth value. And by definition, A conditional $p\to q$ is true whenever $p$ is false, or whenever $q$ is true, or whenever $p$ is false and $q$ is true. There need not be any relationship between $p$ and $q$; we don't need any causal relationship, not even a correlation. Whenever $p$ is false, the implication is true, whether or not q is true or false. Similarly, whenever q is true, the implication is true, regardless of whether p is true or false. $\endgroup$ – Namaste Feb 8 '18 at 17:28
  • $\begingroup$ Note that whenever $p$ from $p\to q$ is false, the statement $p\to q$ is "vacuously true". The conditional/"material" implication $p\to q$ will be true, e.g., if $p$ represents the proposition: "p is a purple pig that flies", whatever q is, true or false, the material conditional/implication is true. $\endgroup$ – Namaste Feb 8 '18 at 17:37
  • $\begingroup$ See also: math.stackexchange.com/questions/439987/… $\endgroup$ – Namaste Feb 8 '18 at 17:48
  • $\begingroup$ For a slightly different approach, see also: math.stackexchange.com/questions/2596766/… $\endgroup$ – Dan Christensen Feb 8 '18 at 21:58
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    $\begingroup$ Consider $p\to q$ as a promise that $q$ will be true if $p$ is true. Well, the only way we can say the promise is not kept is when $p$ is true but $q$ is false. After all, $p\to q$ makes no claim about the value of $q$ when $p$ is false. $$\begin{array}{c:c|l}p & q & p\to q\\ \top & \top & \text{certainly kept, so it is true}\\ \top & \bot & \text{definitely not kept, so it is false}\\ \bot & \top & \text{can not say it is not kept, so it is not false}\\ \bot & \bot & \text{can not say it is not kept, so it is not false} \end{array}$$ $\endgroup$ – Graham Kemp Feb 9 '18 at 2:48