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According to a text "If p and q are mathematical statements (i.e. they can be either true or false but not both) then to show that p or q is true one must consider the following Case 1 : Assume that p is false and show that q must be true Case 2 : Assume that q is false and show that p must be true "

Why do we need to show both the cases to show or is true if we have proved even one is true it is sufficient why does it say to consider both cases. Also can you please explain how we can do it with help of an example. Also what is the meaning of assuming it to be false why do we have to assume anything?

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    If you manage to show one of the cases, then that's enough. However, depending on the statements, it could be difficult. That is when you switch to trying to show the other case. – 2017-02-27
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    Ok so you are saying that the book is trying to say that either one of them is enough prove any one of them according to convenience ? – 2017-02-27
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    Yes, I think that that's what the book is saying. It's what I would say, at least. – 2017-02-27
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    Also what is the meaning of assuming it to be false why do we have to assume anything?? – 2017-02-27
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    Your concern is correct: contraposing Case 2 we have : "Assume that $p$ is false and show that $q$ must be true", that is exactly Case 1. – 2017-02-27
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    In math logic $p \lor q$ is equivalent to $\lnot p \to q$. Thus, to assume $p$ *false* and prove that (with this assumption) $q$ is *true* is enough to assert : $p \lor q$, **if** we are considering "usual" inclusive *or*. – 2017-02-27
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    [Inclusive "or"](https://en.wikipedia.org/wiki/Logical_disjunction#Truth_table) (or *logical disjunction*) means : "at least one of them must be true". This is why the proof of $p \lor q$ amounts to proving $\lnot p \to q$. – 2017-02-27
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    Why cant it be exclusive or? – 2017-02-27
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    It can: but to show that "$p$ (exclusive) or $q$ is true" we need : Case 1) Assume that $p$ is false and show that $q$ must be true; and Case 2) Assume that $p$ is true and show that $q$ must be false. – 2017-02-27
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    See [Exclusive or](https://en.wikipedia.org/wiki/Exclusive_or#Truth_table); we have that $(p \text { XOR } q) \equiv (p \leftrightarrow \lnot q)$. – 2017-02-27
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    "According to a text..." Questions like this can (and should) be improved by giving a citation to the title and author of the text which prompts the post. – 2017-02-28

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You are correct that one does not need to pursue both formulations. Either one is sufficient (and necessary) to give the conclusion "p or q".

You may be concerned about the apparent lack of symmetry shown in either of your (equivalent) formulations Case 1. and Case 2.

Another approach that preserves the underlying symmetry of the roles of $p,q$ in proving $p \lor q$ would be:

Assume $\lnot p$ and $\lnot q$ and derive a contradiction.

In other words, logically $p\lor q$ is equivalent to $\lnot(\lnot p \land \lnot q)$.