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it is given that in csg

if @->#

then length of @ should be less or equal to #

then how aaB->c is a csg???
in Theory of Computation
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2 Answers

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A---> B where |A|<=|B|

So aaB-- >c is not csg.
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This is unrestricted grammar

type-0

The productions can be in the form of α → β where α is a string of terminals and non terminals with at least one non-terminal and α cannot be null. β is a string of terminals and non-terminals.
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