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I run into a bit of confusion when trying to show the equivalence of some statements, for example $$\left(2\le x \lt 5 \right) \vee \left( -1 \lt x \lt 2\right)$$ is equivalent to $$ \left( -1 \le x \lt 5 \right)$$ I can see that it is true that $ \left( -1 \le x \lt 5 \right)$ when either disjunct $\left(2\le x \lt 5 \right)$ or $\left( -1 \lt x \lt 2\right)$ is true and they can't be both true since it would make the equivalent $ \left( -1 \le x \lt 5 \right)$ false.

The part that is unclear is the case when both disjunct are true. They can't be since it would make the equivalent false yet $\phi \vee \psi$ such that $\phi = T$ and $\psi = T$ is a true disjunction. Would that mean that $ \left( -1 \le x \lt 5 \right)$ is equivalent to $\left(2\le x \lt 5 \right)\vee \left( -1 \lt x \lt 2\right)$ is a false proposition?

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    I've never seen a definition for $\lor$ such that two false give a true. – 2017-01-30
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    @kingW3 indeed, thanks for pointing that out, I edited the question to ask it properly. – 2017-01-30

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$(2 \leq x < 5)$ and $(-1 < x < 2)$ cannot both be true because there is no $x$ that satisfies both conditions. If they are both false for some $x$, then $(-1 \leq x < 5)$ is also false, because $x$ is either less than $-1$ or greater than or equal to $5$ to falsify both conditions.

If $\phi$ and $\psi$ are false, then $\phi \vee \psi$ is false too. That's how disjunction is defined. Therefore common sense and propositional logic are in perfect agreement.

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    I edited my question since my interrogation is about when both $\phi$ and $\psi$ are true, which they can't be here, does that mean that the equivalence is no true? – 2017-01-30
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    @NoBoundaries No, it doesn't mean that. It means that in this case you could use either inclusive OR or exclusive OR indifferently. – 2017-01-30