# Inductive Proof of a property of String concatenation

I'm trying to prove that for any 2 strings $\alpha, \beta \in T^\ast$, where $T^\ast$ is the set of all strings on the alphabet $T$, the following holds

$(\alpha \cdot \beta^l) = (\alpha\cdot\beta)^l$

Where for any string $\alpha$:

• $\alpha^l$ denotes the longest prefix
• $\alpha^r$ denotes the shortest suffix
• $|\alpha|$ denotes the length of the string
• $\alpha(n) , n$ a natural number, is the index of the $n$th letter of the string

I'm trying to do it by induction on $|\beta|$, but I'm having a hard time. Here's the base case:

Suppose $|\beta| = 1$

$RHS = (\alpha \cdot \beta )^l$ = $<(\alpha\cdot\beta)(0), ... , (\alpha\cdot\beta)(|\alpha\cdot\beta| - 2)>$ = $<(\alpha \cdot\beta)(0), ..., (\alpha\cdot\beta)(|\alpha|+|\beta| - 2)>$ = $<(\alpha\cdot\beta)(0),...,(\alpha\cdot\beta)(|\alpha|+1-2)>$ = $<(\alpha\cdot\beta)(0), ..., (\alpha\cdot\beta)(|\alpha|-1|)>$

Then for $(\alpha\cdot\beta)(i)$, we have $\alpha(i)$ if $i < |\alpha|$ and $\beta(0)$ if $i = |\alpha|$. I don't think this is right, though.

For my Inductive Case:

Inductive Hypothesis: If $|\beta| = n$, then $(\alpha\cdot\beta^l) = (\alpha\cdot\beta)^l$.

Suppose $|\beta| = n + 1$. Then

$RHS = (\alpha\cdot\beta)^l$ = $<(\alpha\cdot\beta)(0),...,(\alpha\cdot\beta)(|\alpha\cdot\beta|-2)>$ = $<(\alpha\cdot\beta)(0),...,(\alpha\cdot\beta)(|\alpha|+n+1-2)>$ = $<(\alpha\cdot\beta)(0),...,(\alpha\cdot\beta)(|\alpha|+|\beta|-1)>$ = $<(\alpha\cdot\beta(0),...,(\alpha\cdot\beta)(|\alpha\cdot\beta|-1))>$

Which I'm also pretty sure is wrong. I was stuck on this step:

$<(\alpha\cdot\beta)(0),...,(\alpha\cdot\beta)(|\alpha|+n+1-2)>$

For quite a while. I couldn't think of a way to properly apply my inductive hypothesis here.

Is there an easier way to prove this inductively? I feel like I'm missing something that would make this much more intuitive.

• By "longest prefix" to you mean that $\alpha^l$ is $\alpha$ with the last symbol stripped away? What is $\varepsilon^l$? – Henning Makholm Aug 29 '18 at 0:24
• @HenningMakholm Yes, that's correct. $\alpha^l$ would be $\alpha$ with the last symbol stripped away. If $\epsilon$ is the empty string, then $\epsilon^l$ should be $\epsilon$ – enharmonics Aug 29 '18 at 0:56
• x @enharmonics: In that case $\alpha\ne\varepsilon, \beta=\varepsilon$ is a counterexample to what you want to prove. – Henning Makholm Aug 29 '18 at 1:04
• To put Henning Makholm's statement another way: the base case is $|\beta|=0$ not $|\beta|=1$. You can weaken the statement to require that $|\beta|>0$ if you want. – Derek Elkins Aug 29 '18 at 2:02
• Probably not what you're looking for, but there's a very slick way of proving such statements with initiality, outlined here: homepages.cwi.nl/~janr/papers/files-of-papers/… – Rafay Ashary Aug 29 '18 at 3:25

## 2 Answers

Don't take it the wrong way, but the main reason for which you don't find an intuitive proof is your poor notation. So let me first reformulate your question with a different notation (mostly taken from Lothaire's book Combinatorics on words).

Let $A$ be an alphabet and let $A^*$ be the set of all words on $A$. Note that $A^*$ is a monoid for the concatenation product. The identity of this product is the empty word $1$.

According to your definition the longest prefix $LP(a_1 \dotsm a_n)$ of $a_1 \dotsm a_n$ is $a_1 \dotsm a_{n-1}$ if $n > 0$ and is $1$ if $n = 0$. I claim that for all words $u, v$, one has $uLP(v) = LP(uv)$. The result is trivial if $v = 1$. Suppose that $v \not= 1$. Setting $u = a_1 \dotsm a_n$ and $v = b_1 \dotsm b_m$ (with $m > 0$), one gets $$uLP(v) = a_1 \dotsm a_nb_1 \dotsm b_{m-1} = LP(uv)$$ This is a very intuitive proof that does not require any induction. The proof for the suffixes is similar.

I like your thinking on this. We should use induction.

Suppose $\alpha, \beta$ are both $\epsilon$ the empty string. Then clearly it holds since $\epsilon \epsilon^l = \epsilon = (\epsilon \epsilon)^l$. Now assume that it's true for $|\alpha| = m, |\beta| = n$. Then if $|\alpha'| = n + 1$ then we can write $\alpha' = a\alpha$ for some $a\in \Sigma$ our alphabet. Then $\alpha' \beta^l = a (\alpha \beta^l) = a(\alpha \beta)^l = (a\alpha\beta)^l$. Where we need a lemma:

If $|\alpha| = 1$ then $\alpha \beta^l = (\alpha \beta)^l$ for any string $\beta \in \Sigma^*$. It's clearly true for $\beta = \epsilon$ since $\alpha \epsilon^l = \alpha \epsilon \neq (\alpha \epsilon)^l$ clearly that doesn't hold in general. So we're in trouble. Let's assume that $|\beta| \gt 0$. Then $\alpha b^l = \alpha \epsilon = (\alpha b)^l$. So it's true in the base case of $|\beta| = 1$. Now assume true for $|\beta| = r$. Then if $|\beta'| = r + 1$ we can write $\beta' = \beta b$ for some $b \in \Sigma$. So that we have: $$\alpha \beta'^l = \alpha \beta = (\alpha \beta b)^l = (\alpha \beta')^l$$.

So, as you can see it can be proven by two separate inductions. One for $n$ and one for $m$. I'll let you prove the $m$ one.