HW 6.2-solution (Continued): 10) DeMorgan's law: (x+y)' = x'y (xy)=x+yVx, y EB Proof By duality we...

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HW 6.2-solution (Continued): 10) DeMorgan's law: (x+y)' = x'y (xy)=x+yVx, y EB Proof By duality we need to prove only the first equality: (x + y)' = x'y' We use the same reasoning as for the previous property, 9). That is, we use the axiom that defines the complement, namely 5): 5) VxEB, 3 x' EB, x + x¹ = 1 and xx' = 0 In the first equality, which is the element and which do we have to prove is the complement? Element = x+y, and its complement = x'y'. Substitute in 5). We need to prove: x+y + xy = 1 (x + y). (x' y) = 0 HW 6.2-solution (Continued): 10) DeMorgan's law: (x+y)' = x'y (xy)=x+yVx, y EB Proof By duality we need to prove only the first equality: (x + y)' = x'y' We use the same reasoning as for the previous property, 9). That is, we use the axiom that defines the complement, namely 5): 5) VxEB, 3 x' EB, x + x¹ = 1 and xx' = 0 In the first equality, which is the element and which do we have to prove is the complement? Element = x+y, and its complement = x'y'. Substitute in 5). We need to prove: x+y + xy = 1 (x + y). (x' y) = 0