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@@ -69,8 +69,8 @@ $$ e_N = (a_1\cup (b_2b_3))^*(c_4^+ \cup (a_5 \cup (b_6b_7)))$$
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3. Isolate the subexpressions and put them into evidence
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$$ (a_1\cup (b_2b_3))^*\;\;\;\;c_4^+ \cup (a_5 \cup (b_6b_7))$$
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$$ a_1\cup (b_2b_3)\;\;\;\;\;\;c_4^+\;\;\;\;a_5 \cup (b_6b_7)$$
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-$$ a_1\;\;\;\;b_2b_3\;\;\;\;\;\;c_4^\;\;\;\;a_5 \;\;(b_6b_7)$$
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-$$ a_1\;\;\;\;b_2\;\;\;b_3\;\;\;\;\;\;c_4^\;\;\;\;a_5 \;\;\;b_6\;\;\;b_7$$
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+$$ a_1\;\;\;\;b_2\;\;b_3\;\;\;\;\;\;c_4^+\;\;\;\;a_5 \;\;(b_6b_7)$$
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+$$ a_1\;\;\;\;b_2\;\;\;b_3\;\;\;\;\;\;c_4^+\;\;\;\;a_5 \;\;\;b_6\;\;\;b_7$$
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<!--*-->
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### Different parenthesization
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@@ -89,8 +89,8 @@ If the two regexp $e'$ and $e''$ can be factored as
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$$e'=\alpha\beta\gamma \;\;\; e''=\alpha\delta\gamma$$
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With
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- $\beta$ subexpression of $e'$
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-- $\gamma$ subexpression of $e''$
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-- $\gamma$ is a *choice* of $\beta$
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+- $\delta$ subexpression of $e''$
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+- $\delta$ is a *choice* of $\beta$
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##### examples
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- one step derivation
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@@ -174,6 +174,10 @@ An __element__ can be a __terminal__ (alphabetic symbol) or a __compound object_
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##### example:
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List with separators and start-marker of end-marker
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$$ie(se)^*f\;\;\;i[e(se)^*]f$$
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+i: inizio
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+e: elemento
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+s: separatore
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+f:fine
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### Substitution
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Operation that replaces the terminal characters of the source language with the phrases of the destination languages
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