5. Let E= {0, 1,*,=} and MUL = {x*y=z | x, y, z are binary integers,...

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5. Let E= {0, 1,*,=} and MUL = {x*y=z | x, y, z are binary integers, and z is the product of x and y}. For example, string "11*101-1111" is in MUL, while string "11*101=10" is not in MUL Using Pumping Lemma prove that MUL is not a regular language. 5. Let E= {0, 1,*,=} and MUL = {x*y=z | x, y, z are binary integers, and z is the product of x and y}. For example, string "11*101-1111" is in MUL, while string "11*101=10" is not in MUL Using Pumping Lemma prove that MUL is not a regular language.

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To prove that MUL is not a regular language using the Pumping Lemmawe assume that MUL is regularThis means that there exists a finite automaton M that accepts all strings in MUL and rejects all string... View the full answer