Assume that we would like to simplify the following Context Free Grammar, G: G = (V,T,S,P) where V={S,a,b,c,d} is the set of non-terminal symbols, T={0,1,2} is the set of terminal symbols, S is the start symbol and P={S→ac\d, a→0, c→1|bc, d→Ob} is the set of the production rules. In our first step, we start by the terminal symbol set, T, and we find set W1 whose members yield to WO=T, then we find W2 whose members yield to W1, etc. At the end of this step, our grammar will be simplified as following: The terminal set is {0,1,2}, and the non-terminal set is S, a, c the start symbol is S, and the set of production rules is {S→ ac , а -0, с— 1|bc In the next step, we begin by the start symbol, S: Z0=S, then we find the set Z1 whose members are yielded from Z0, then we find Z2 whose members are yielded from Z1, etc. At the end of this step, our grammar will be simplified as following: The terminal set is 0, 1 and the non-terminal set is S, a , c the start symbol is S, and the set of production rules is {S→ ac а -0, с— 1 }.

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Assume that we would like to simplify the following Context Free Grammar, G:
G = (V,T,S,P) where V={S,a,b,c,d} is the set of non-terminal symbols, T={0,1,2} is the set of terminal symbols, S is the start symbol and P={S→ac|d, a→0,
c-1|bc, d→Ob} is the set of the production rules.
In our first step, we start by the terminal symbol set, T, and we find set W1 whose members yield to WO=T, then we find W2 whose members yield to W1, etc.
At the end of this step, our grammar will be simplified as following:
The terminal set is {0,1,2}, and the non-terminal set is S, a, c
the start symbol is S, and the set of production rules is
{S→ ac
, а-0, с— 1|bс
}
In the next step, we begin by the start symbol, S: Z0=S, then we find the set Z1 whose members are yielded from Z0, then we find Z2 whose members are
yielded from Z1, etc.
At the end of this step, our grammar will be simplified as following:
The terminal set is 0, 1
, and the non-terminal set is S, a , c
, the start symbol is S, and the set of production rules is
{S→ ac
, а--0, с— 1
}.
Transcribed Image Text:Assume that we would like to simplify the following Context Free Grammar, G: G = (V,T,S,P) where V={S,a,b,c,d} is the set of non-terminal symbols, T={0,1,2} is the set of terminal symbols, S is the start symbol and P={S→ac|d, a→0, c-1|bc, d→Ob} is the set of the production rules. In our first step, we start by the terminal symbol set, T, and we find set W1 whose members yield to WO=T, then we find W2 whose members yield to W1, etc. At the end of this step, our grammar will be simplified as following: The terminal set is {0,1,2}, and the non-terminal set is S, a, c the start symbol is S, and the set of production rules is {S→ ac , а-0, с— 1|bс } In the next step, we begin by the start symbol, S: Z0=S, then we find the set Z1 whose members are yielded from Z0, then we find Z2 whose members are yielded from Z1, etc. At the end of this step, our grammar will be simplified as following: The terminal set is 0, 1 , and the non-terminal set is S, a , c , the start symbol is S, and the set of production rules is {S→ ac , а--0, с— 1 }.
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