Propositional Calculus: Stating and verifying readability and unique readability of a given language $\mathcal{L^*}$ Problem: Consider the set of symbols * and #. Let $\mathcal{L^*}$ be the smallest set $L$ of sequences of these symbols with the following properties:
a) The length one sequences $\langle\mbox{*}\rangle$ and $\langle$#$\rangle$ belong to $L$
b) If $\sigma$ and $\tau$ belong to $L$, then so do $\langle\mbox{*}\rangle+\sigma+\langle\mbox{#}\rangle$ and $\langle\mbox{*}\rangle+\sigma+\tau+\langle\mbox{#}\rangle$.
State Readability and Unique Readability for $\mathcal{L^*}$ and determine for each whether it holds.
How do we determine what makes it readable? How do we construct these criteria?
 A: Unique Readability for $\mathcal{L^*}$: Suppose $\phi \in \mathcal{L^*}$. Then exactly one of the following conditions applies :
(1) $\phi =\langle \mbox{*} \rangle$
(2) $\phi = \langle \mbox{#} \rangle$
(3) There is a $\psi \in \mathcal{L^*}$ such that $\phi = \langle \mbox{*} \rangle + \psi + \langle \mbox{#} \rangle $.
(4) There are $\psi_1, \psi_2 \in \mathcal{L^*}$ such that $\phi = \langle \mbox{*} \rangle + \psi_1 + \psi_2 + \langle \mbox{#} \rangle $. 
Further, $\psi$ in (3) and $\psi_1, \psi_2$ in (4) are unique.
A quick check shows that the subset $L$ of $\mathcal{L^*}$ satisfying the above clauses is $\mathcal{L^*}$ itself. We want to show that readability holds, meaning that the cases are mutually exclusive.
Readable?: Say $\phi = \langle \mbox{*} \rangle$ Then it obviously cannot be (2), and vice versa. It cannot be (3) since (3) must be a sequence of at least three (since the empty set is not a "formula"). It cannot be (4) either since (4) must be a sequence of length at least four. The last two results can also be said for  $\phi = \langle \mbox{#} \rangle$. All that remains is to show that (3) and (4) are mutually exclusive. Suppose $\phi = \langle \mbox{*} \rangle + \psi + \langle \mbox{#} \rangle = \langle \mbox{*} \rangle + \psi_1 + \psi_2 + \langle \mbox{#} \rangle $.  
$\implies \psi = \psi_1 + \psi_2$. Can we make this happen? (Can the concatenation of two formulas be a formula?)
Take $\psi_1 = \mbox{*} \tau \mbox{#} $ and $\psi_2= \mbox{*} \sigma \mbox{#}$. Call $\psi =\psi_1 +\psi_2 = \mbox{*} \tau \mbox{#}\mbox{*} \sigma \mbox{#}$ is an element if $\tau \mbox{#}\mbox{*} \sigma$ is. Suppose $\tau = \mbox{*}, \sigma = \mbox{#}$. Then we have $\mbox{*} \mbox{#} \mbox{*}\mbox{#}$ is an element by (4), so $\mbox{*} \mbox{*} \mbox{#}\mbox{*} \mbox{#} \mbox{#}$ is of form (3) and (4) and thus they are not mutually exclusive $\therefore$ $\mathcal{L^*}$ is not readable.
Uniquely Readable?: Unnecessary since the language is not readable.
A: I think you have the statement for unique readability written out right, but the problem is that the language is not uniquely readable.
To see this, we can construct
a=<>+<>+<#>,
b=<>+<#>+<#>,
c=<>+<>+<#>+<>+<#>+<#>
Now, since a, b, and c are all in L*, <>+a+b+<#> and <>+c+<#> are also each in L*.
But <>+a+b+<#> = <>+c+<#>, so a+b=c, which is a contradiction. 
