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L1 = {a,b), L2 = (c,d). Then L1cdot L2 = {ac, ad, bc, bd) cdot = concatenation True False Then L 1 Xcdot L2 (ac.

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L1 = {a,b), L2 = (c,d). Then L1\cdot L2 = {ac, ad, bc, bd) \cdot = concatenation True False

Then L 1 Xcdot L2 (ac. ad. bc- bd) \cdot concatenation True False previous

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