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Let E={aibj|i≠jand2i≠j}. Show thatE is context-free language.

Short Answer

Expert verified

It is shown thatE is a context-free language.

Step by step solution

01

Define context-free language

The context-free language is generated by context-free grammar. These languages are accepted by Pushdown Automata. These are the supersets of regular languages.

Consider context-free languages L1described as G1=(V1,S,R1,S1).

Consider context-free languageL2 described asG2=(V2,S,R2,S2).

02

Explain the solution

Split it as unions of three languages.

aibj|i>j̀Eaibj|i<jand2i>j∪aibj|2i<jG1=aibj|i>jG2=aibj|i<jand2i>jG3=aibj|2i<j

G1andG3arequiteeasytofigureout.forG1S→aSb|aS1S1→aS1|ε


for G3

S→aSbb|S1bS1→S1b|εNowtrickypartisG2.

S→aSb|aS1bS1→aS1bb|aS2bbS2→ε

It is proved very clearly, that Swillgenerateanamb2mbnthatisequivalenttoan+mbn+2m,

wherem,nmustbegreaterthan0.

Hence,E is a context-free language.

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