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Prove that if the first symbol of $\alpha$ is a wff of a First order language, then there is a unique pos. integer i, and a unique wff $\beta$, such that $\alpha$ is $\forall v_i \beta$.

I am reasonably confident of my proof for $\alpha$ is of the form $\forall v_i \beta$ but I'm lost on how to actually prove that this i is unique and $\beta$ is unique. Maybe it could start with assume $\alpha = \forall v_i \beta_1 = \forall v_j \beta_2$ where $i \neq j$ and $\beta_1 \neq \beta_2$ but I don't know how to proceed from this.

Could someone give me some help on this one. I could only find answers to unique readability of sentential logic.

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I suspect you're thinking about this a bit too hard.

If $i\not=j$, then the strings "$\forall v_i\beta_1$" and "$\forall v_j\beta_2$" are clearly different (their second terms are different); similarly, if $\beta_1\not=\beta_2$, then let $n$ be the least place at which the strings $\beta_1$ and $\beta_2$ differ - clearly the strings "$\forall v_i\beta_1$" and "$\forall v_j\beta_2$" have different $(n+2)$th terms.


To give a bit of an overall picture, which might be helpful:

The quantifier case is rather trivial, since quantifiers can just be "peeled off." The Boolean combinations are really where things get complicated; essentially, the problem is that having multiple Boolean operations might let you "split" a given wff in different contradicting ways: e.g. think about $$p\vee q\wedge r,$$ which could be read as "$p$ OR ($q$ AND $r$)" or "($p$ OR $q$) AND $r$." This is an example of unique readability failing for a certain class of expressions; the hope is that this sort of nonsense will be prevented by the way parentheses are used in the definition of a well-formed formula

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