# proving something is not a well formed formula

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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.

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 truth-functional 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 first-order 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, 2022

• 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 years
This depends heavily on the particular definition of well-formed formula.
• Mauro ALLEGRANZA over 8 years
I 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 years
I 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 $A_i$?