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Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
First model: Computer has a ?xed number of bits of storage. You will model this by limiting your program to a single ?xed-precision unsigned integer variable, e.g., a single one-by
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Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B no
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