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Explain the context free language closure under union operation?
If L1 and L2 are CFLs, then their union L1 + L2 is a CFL.Here CFL refers to Context Free Language.
L1 CFL implies that L1 has a CFG, let it is CFG1, that generates it.
Assume that the nonterminals in CFG1 are S, A, B, C, . . ..
Change the nonterminal in CFG1 to S1, A1, B1, C1, . . ..
Don’t change the terminals in the CFG1.
L2 CFL implies that L2 has a CFG, Let it is CFG2, that generates it.
Assume that the nonterminals in CFG2 are S, A, B, C, . . ..
Change the nonterminal in CFG2 to S2, A2, B2, C2, . . ..
Don’t change the terminals in the CFG2.
Now CFG1 and CFG2 have non intersecting sets of nonterminals.
We create a CFG for L1 + L2 as follows −
Include all of the nonterminal S1, A1, B1, C1, . . . and S2, A2, B2, C2, . . ..
Include all the productions of CFG1 and CFG2.
Create a new nonterminal S and a production.
S → S1 | S2
CFG1 for L1 S → S | Aa | Bb | ∧ A → Sa | bB | ab B → S | a CFG2 for L2 S → aS | aA | Bb | ∧ A → aS |bB| ab B → Ba | b
To construct CFG for L1 + L2, follow the below mentioned steps −
Transform CFG1 as follows −
S1 → S1| A1a | B1b | ∧ A1 → S1a | bB1 | ab B1 → S1 | a
Transform CFG2 as follows
S2 → aS2 | aA2 | B2b | ∧ A2 → aS2 |bB2| ab B2 → B2a | b
Construct CFG for L1 + L2 as follows −
S → S1 | S2 S1 → S1| A1a | B1b | ∧ A1 → S1a | bB1 | ab B1 → S1 | a S2 → aS2 | aA2 | B2b | ∧ A2 → aS2 |bB2| ab B2 → B2a | b
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