If $\theta$ is a WFF, then $((\theta))$ is not a WFF

Question

I can show that if $\theta$ is a WFF, then $(\theta)$ is not a WFF. Suppose $ (\theta )$ is a WFF. Then by unique readability, we know $(\theta) \equiv \neg\alpha$ or $(\theta) \equiv (\alpha \square \beta)$ for some formulae $\alpha$ and $\beta$ and $\square$ being one of the four binary connectives. And as $\neg$ is not the same as (, $(\theta) \equiv (\alpha \square \beta)$, So $\theta \equiv \alpha \square \beta$. So, $\alpha$ is a proper initial segment of $\theta$, a contradiction as $\alpha$ is a formula. Hence $(\theta)$ is not a WFF. But I am having difficulty showing $(( \theta ))$ is not a WFF. Any hints? And how to show the general case of $n$ parenthesis on either side?

Answer 1

Given some wff $\theta$, if $((\theta))$ was wff then $(\theta)$ must look like $\alpha \square \beta$ for some wffs $\alpha$, $\beta$. Now, since $\alpha$ is a wff that must start with $($ we can write that $(\theta)=\alpha\square \beta=(\gamma\square \phi)\square \beta$ for some wffs $\gamma,\phi$ but that means $\gamma$ is a proper initial segment of $\theta$, a contradiction. For the general case repeatedly use the same idea.

Answer 2

For the general case, here is the induction:

Let $t_n$ be a string of $n$ "(" symbols and $v_n$ be a string of $n$ ")" symbols. We claim that if $s$ is a string and $t_nsv_n=\phi_1 \square \phi_2$ for formula $\phi_1,\phi_2$, then $\phi_1 \square \phi_2=t_n\theta_1 \square_1 \theta_2)\square_2 \theta_3)... )\square_{n+1} \theta_{n+2}$ for some formulae $\theta_1,\theta_2..\theta_{n+2}$ and connectives $\square_1,\square_2,..\square_{n+1}$* and we show by induction on $n$:

For the base case $(n=0)$ just put $\theta_1=\phi_1, \theta_2=\phi_2$ and $\square=\square_1$.
For the inductive step, suppose $s$ is a string and $t_{n+1}sv_{n+1}=\phi_1 \square \phi_2$. Then $t_n(s)v_n=\phi_1 \square \phi_2$ and so by our IH, $\phi_1 \square \phi_2=t_n\gamma \square_2 \theta_3)\square_3 \theta_4)... )\square_{n+2} \theta_{n+3}$ for some formulae $\gamma, \theta_3,.. \theta_{n+3}$ and connectives $\square_2, \square_3, ..\square_{n+2}$. But then $\gamma$ starts with ( and so by unique readability, is of the form $(\theta_1 \square_1 \theta_2)$ for some formula $\theta_1,\theta_2$ and connective $\square_1$. Hence $\phi_1 \square \phi_2=t_{n+1}\theta_1 \square_1 \theta_2)\square_2 \theta_3)... )\square_{n+2} \theta_{n+3}$, the required format.

Now suppose $\theta$ is a formula and $t_n \theta v_n$ is a formula. Then by unique readability, $t_n \theta v_n=(\phi_1 \square \phi_2)$ for some formulae $\phi_1,\phi_2$ and connective $\square$. And so, $t_{n-1} \theta v_{n-1}=\phi_1 \square \phi_2$. Hence by *, $t_{n-1} \theta v_{n-1}=t_{n-1}\theta_1 \square_1 \theta_2) \square_2 \theta _3)..)\square_n \theta_{n+1}$ for some formulae $\theta_1, \theta_2,..\theta_{n+1}$ and connectives $\square_1,\square_2,..\square_n$. But then $\theta_1$ is a proper initial segment of $\theta$, a contradiction. Hence $t_n \theta v_n$ is not a formula