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L1={$a^nb^m$:n>=0,m>n}

L2={$a^nb^{2n}$:n>=0}

find the grammar for L1L2
asked in Theory of Computation by Active (4k points)
edited by | 39 views

1 Answer

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Best answer
S -> AB

A ->aAC | C

C -> bC | b

B -> aBbb | $\varepsilon$
answered by Active (3.6k points)
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