# For any sentence $A$, the only sentence which is an initial segment of $A$ is $A$ itself

I'm having trouble following this proof about the syntax of first-order logic. When $$A$$ and $$B$$ are expressions, we say that $$B$$ is an initial segment of $$A$$ if and only if $$A$$ can be obtained by concatenating some (possibly no) symbols to the (right-hand) end of $$B$$. For example:

$$\neg$$

is an initial segment of $$\neg \neg A$$

Now the inductive hypothesis is that for some arbitrary $$A$$ every sentence shorter than $$A$$ has this property. Then we reason by cases. The atomic case is straightforward, but the proof lost me on the case for negation. It says:

Say that $$A$$ is $$\neg B$$ for some sentence $$B$$. Let $$A^*$$ be any sentence which is an initial segment of $$A$$. So $$A^*$$ is $$\neg B^*$$, for some sentence $$B^*$$. Evidently, $$B^*$$ is an initial segment of $$B$$. But $$B$$ and $$B^*$$ are both sentences that are shorter than $$A$$; so by the induction hypothesis, $$B^*$$ must just be $$B$$ itself. Hence $$A^*$$ is $$A$$ itself, and so $$A$$ has the induction property.

This proof-segment is quite quick and compressed, and I'm not seeing the moves that it presents as obvious.

First of all:

Why is $$B^*$$ 'evidently' an initial segment of $$B$$? After all, $$B$$ and $$B^*$$ could be anything, string of conjunctions, conditionals, whatever? The author treats this as an obvious step, but I don't see it.

Second:

How could $$B^*$$ not be $$B$$ when we have said that $$A$$ is $$\neg B$$ and $$\neg B^*$$ is an initial segment of $$A$$?

Any help would be much appreciated.

• We have that $A$ is $(\lnot B)$ and we assume that we can find $A^*$ that is an initial segment of $A$; this means that $A$ must be $(\lnot \text { something})$ and we call that something $B^*$. Obviously $B^*$ must be an initial segment of $B$: try with an example: $A$ is $\lnot (C \land D)$ and thus $B$ is $(C \land D)$ and thus $B^*$ must be this way: $(C \ldots$. Nov 1, 2022 at 19:20
• If $B^*$ is an initial segment of $B$ we apply the induction hypotheses that holds because $B$ is shorter than $A$ and thus $B^*$ is equal to $B$ itself by assumption. Nov 1, 2022 at 19:24
• sadly I don't think I follow with the first one Nov 1, 2022 at 21:05