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I'm trying to answer the following exercise question:

Exercise 1. Let $Σ =\{0, 1\}$ and let $L$ be a set of strings. As seen in lectures, $L$ induces an equivalence relation denoted $≡_L$ over the set of strings. Prove $≡_L$ is an equivalence relation, i.e. it satisfies the 3 properties (symmetry, reflexivity and transitivity).

As I understand it there is no equivalence relation described in the question. An equivalence relation would be something like "$a$ is related to $b$ (i.e $a ≡_L b$) if $a-b$ is an integer".

So here is my attempt at transcribing the relevant lecture notes:

$L$ is a set
$L-x=\{y : xy ∈ L\}$
$x ≡_L y \iff L-x = L-y$

I am unclear what is being described here. "$L$ is a set" - ok, in this $L$ could be say $\{0, 01, 0101\}$ or similar. The following lines describe what the relation actually is but I can't understand them. Can anyone please help me so I can go about proving the transitivity/reflexivity/symmetry of the equivalence to answer the question.

Is it saying the set $L-x=y$, such that $xy$ is an element of the set $L$? That doesn't seem intuitive.

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  • $\begingroup$ $L$ is a set of strings, and for any single string $x$, $L-x$ denotes the set of suffixes for which $x$ is a prefix. That is to say, given $x$, collect all the strings in $L$ that start with $x$. The set of strings that you get when you remove the initial $x$ in that collection—that's what $L-x$ is. For instance, if $L$ is, say, all strings of length $6$ that have an even number of $1$s in them, then $L-001 = \{001, 010, 100, 111\}$, because the set of strings in $L$ that begin with $001$ is $\{001001, 001010, 001100, 001111\}$. Remove the initial $001$ from each, and you get $L-001$. $\endgroup$ – Brian Tung Mar 11 at 3:20
  • $\begingroup$ You have to show, presumably, that the relation $x \sim_L y \Leftrightarrow L-x = L-y$ is in fact an equivalence relation, whatever $L$ happens to be. If you think about what this relation is, this shouldn't be too difficult, I don't think. $\endgroup$ – Brian Tung Mar 11 at 3:22
  • $\begingroup$ Thank you that seems right on the money, the question makes sense now. So ≡L is basically "the prefix and suffix are the same" ex {010010, 101101}? p.s I can't seem to upvote or select as the correct answer (Im normally on stackoverflow and cs.stackexchange) I assume the community will do this? $\endgroup$ – Jay.F Mar 11 at 3:31
  • $\begingroup$ Are you talking about the equivalence relations in the context of the Myhill-Nerode theorem? $\endgroup$ – ShyPerson Mar 11 at 4:45
  • $\begingroup$ Myhill-Nerode theorem is the theme of the overall assignment $\endgroup$ – Jay.F Mar 11 at 4:56
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The wording of the exercise could easily have been made more precise. I suppose that $L$ is a set of strings in $\Sigma,$ that is, $L \subseteq \Sigma,$ and that when we say $L$ "induces an equivalence relation ... over the set of strings," we really mean an equivalence relation over $\Sigma.$

If $L$ is a set of strings and $x$ is a string, then the definition from the lecture says that $L - x$ also is a set of strings, specifically all the strings that you can append to $x$ in order to get a string in $L.$

You could think of it this way: a string $y$ is a "good" suffix of $x$ if you can append $y$ to $x$ and get a string in $L,$ that is, $xy\in L.$ Then $L - x$ as the set of "good" suffixes of $x.$ Note that if $x$ itself is in $L,$ then one of the "good" suffixes is the empty string.

The relationship according to your notes is that $x \equiv_L y$ iff the "good" suffixes of $x$ are exactly the same as the "good" suffixes of $y.$

If $L = \{0, 01, 0101\},$ as in your example, then $L - 010 = \{1\},$ $L - 01 = \{e, 01\}$ (where $e$ is the empty string), and $L - 0 = \{e, 1, 101\}.$ But $L - 1 = \emptyset$; if you start a string with $1,$ there is no way to finish the string in order to make it something in $L.$ On the other hand, $L - e = L.$ (Actually that last fact would be true no matter what you chose for the members of $L.$)

So $01 \not\equiv_L 010$ in your example, because $\{e, 01\} \neq \{1\}.$ Likewise $0 \not\equiv_L 01.$ But $1 \equiv_L 11,$ because $L - 1 = L - 11 = \emptyset.$ In fact if you choose any $x$ and $y$ that are not prefixes of any string in $L$ then $L - x = L - y = \emptyset$ and therefore $x \equiv_L y.$

As for proving the equivalence relation, it's a relatively mechanical task. You need to show that if $x, y, z \in \Sigma,$ then $x\equiv_L y$ implies $y\equiv_L x$ (symmetry), $x\equiv_L x$ (reflexivity), and if $x\equiv_L y$ and $y\equiv_L z$ then $x\equiv_L z$ (transitivity). For example, for symmetry, if the set of "good" suffixes of $x$ is the same as the set of "good" suffixes of $y,$ is it always true that the set of "good" suffixes of $y$ is the same as the set of "good" suffixes of $x$? The trick is to write this formally, using the symbology from your notes.

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