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Look at below formulas:

$(\forall a)(a<b)$: $a$ is bound and $b$ is free. $(\forall b)(b<a)$: $b$ is bound and $a$ is free.

Now if one considers $[(\forall a)(a<b)$ or $(\forall b)(b<a)]$ all we can say is that: 1st occurrence of $b$ and 2nd occurrence of $a$ are free, whereas 1st occurrence of $a$ and 2nd occurrence of $b$ are bound.

Can we claim anything about overall occurrence of $a$ and $b$ to be bound or free?

In other words, is $[(\forall a)(a<b)$ or $(\forall b)(b<a)]$ a sentence?!

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    It is not a sentence because $b$ is free in the left disjunct and $a$ is free in the right one. – Mauro ALLEGRANZA Jan 15 '17 at 21:32
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    Although your $[(\forall a)(a<b)$ or $(\forall b)(b<a)]$ is a legal expression I would argue it is terrible style because of just the confusion you are asking about. Using the same letter as two different variables makes it difficult to read. You want to make things easy on the reader, not difficult, and writing it as $[(\forall a)(a<c)$ or $(\forall b)(b<d)]$ makes it much clearer what is going on. – Ross Millikan Jan 15 '17 at 21:40
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    As you correctly write, it is an occurrence of a variable that is free or bound, and not the variable. – Mauro ALLEGRANZA Jan 15 '17 at 21:55
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    @MauroALLEGRANZA: Not for some authors; see my answer. =) – user21820 Jan 17 '17 at 11:59
  • @user21820 - agreed... but we can match the two "points of view" : a variable $x$ is free in a formula $\varphi$ if it has one or more free occurrences; otherwise, it is bound. – Mauro ALLEGRANZA Jan 17 '17 at 13:13
  • @MauroALLEGRANZA: Yup certainly, though I would refrain from using the word "bound" since a variable might be not free might simply because it does not occur in the formula at all. – user21820 Jan 17 '17 at 14:11

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Some textbooks (such as Rautenberg's linked from this post) do define the set of free variables of a formula, such that it include variables that occur at least once in the formula in an unbound state. In particular $free(\ \forall a(a<b) \lor \forall b(b<a)\ ) = \{a,b\}$. Of course, once you fix the definition of "free variables", you have to stick to it very carefully from then on, and the details are very important when it comes to the deductive rules (in Rautenberg's case regarding "collision-free substitutions").

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