Question: Let G = ({S, A, B, C, D}, sigma, R, S) be the context-free grammar with sigma ={a, b} and rules R = S rightarrow

Let G = ({S, A, B, C, D}, sigma, R, S)

Let G = ({S, A, B, C, D}, sigma, R, S) be the context-free grammar with sigma ={a, b} and rules R = S rightarrow AD | bbC | SaBS A rightarrow BAB | elementof B rightarrow SB | b C rightarrow ACD | BA | DAS D rightarrow BaaC | elementof. (Since the period is not in sigma, it is just punctuation.) Find a grammar G' without elementof-rules such that L(G') = L(G) \ {elementof}. Show clearly which variables in G are nullable. Is elementof elementof L(G)? It is OK if your final G' looks "gross" so long as you show the steps of the algorithm clearly. Let G = ({S, A, B, C, D}, sigma, R, S) be the context-free grammar with sigma ={a, b} and rules R = S rightarrow AD | bbC | SaBS A rightarrow BAB | elementof B rightarrow SB | b C rightarrow ACD | BA | DAS D rightarrow BaaC | elementof. (Since the period is not in sigma, it is just punctuation.) Find a grammar G' without elementof-rules such that L(G') = L(G) \ {elementof}. Show clearly which variables in G are nullable. Is elementof elementof L(G)? It is OK if your final G' looks "gross" so long as you show the steps of the algorithm clearly

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