proving something is not a well formed formula
The problem is that the concep of well formed formula depends on the language you are working with.
A language has a syntax, i.e. rules that allows you to "build" terms (i.e."names") and expressions (i.e."sentences").
If I assume that you are working in formalized arithmetic, usualy variable like "x" are terms standing for numbers; so, the expression $x > y$ is well formed. But, in this case $x$ and $y$ are not wff.
If instead we are working with the language of propositional logic, where the variables ($x$ and $y$) are called sentential variables, the expression $x > y$ is not well formed.
Again, we need to know what language are you handling.
About concatenation:
but the comment about concatenation of two wffs not being allowed is really the crux of my issue.
If I work with propositional logic, the language is made by sentential variables, like $p, q, r, ...$ and truthfunctional connectives, $\lnot$ (negation, unary), $\land$ (conjunction, binary), $\lor$ (disjunction, binary), $\rightarrow$ (conditional, binary).
The rules available to build expressions are (tipically) :
(i) every sentential variable is a wff
(ii) if $A$ ia a wff, then $\lnot A$ is a wff
(iii) if $A$ and $B$ are wffs, then $A \lor B$, $A \land B$ and $A \rightarrow B$ are wffs
(iv) nothing else is a wff.
Reviewing the above rules, you can see that there are no rules allowing you to juxtapose two wff without interposing a connective; e.g. :
$AB$ and $\lor A$ are not wffs.
If you are working in firstorder number theory, you have individual variables, like $x$ and $y$, you have the above connectives, you have the equality symbol and the quantifiers, (for simplicity, we assume that also $>$ is a primitive symbol).
The following are examples of wffs :
$\forall x (x > 0)$ and $x = 0$
Again, you have rules like the above, that do not allow you to "build" an expression like :
$(x = 0) \forall x (x > 0)$
But there are languages ($\lambda$calculus) where concatenation is allowed; and then there are programming languages ...
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Sandie H.
Updated on August 11, 2022Comments

Sandie H. about 17 hours
Prove if $W$ is well formed formula (wff) them $W$ cannot be written as $W_1W_2$ where $W_1$ and $W_2$ are wffs
$$\text{wff} :=\begin{cases} (A_i)\\ (\sim x)&\text{if } x \text{ is a wff}\\ (x>y)&\text{if }x, y \text{ are wffs}\end{cases}$$
I apologize if I am not expressing myself well, but the comment about concatenation of 2 wffs not being allowed is really the crux of my issue. Why isn't this allowed? By the laws of a legal parenthesis string (lps) this is acceptable as long as what ever is inside of the two parentheses are lps themselves. Isn't a wff an lps in it's own right? Again, I am sorry, I'm sure this is probably simple but I always get caught up in the semantics of thing and want to make sure I am seeing things correctly.

Dustan Levenstein over 8 yearsThis depends heavily on the particular definition of wellformed formula.

Mauro ALLEGRANZA over 8 yearsI rephrase the comment by Dunstan : if you want a satisfying answer, please ask an understandable question. If you are working in propositional logic, a plausible reading of your statement has a definite answer; something like : the operation of concatenation of formulas is not allowed; between two wffs you must interpose a binary connective, like : $\lor$, $\land$ or $\rightarrow$.

Hagen von Eitzen over 8 yearsI tried to format the question towards legibility, but I'm not sure what to make of the last part. Also, what can be said about the $A_i$?
