When are the inputs to a NAND gate, according to De Morgan's theorem, the output expression could be:-A-X = A + BB-C-X = (A)(B)D--A-X = A + B

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When  are the inputs to a NAND gate, according to De Morgan's theorem, the output expression could be:-A-X = A + BB-C-X = (A)(B)D--A-X = A + B

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The correct option is  A X = A + B A AND B = A*B A NAND B = BAR(A*B) BAR(A) NAND BAR(B) = BAR[BAR(A)* BAR(B)] According to DeMorgan&#39;s Theorem: BAR[BAR(A)* BAR(B)] = BAR[BAR(A)] + BAR[BAR(B)] According to Double negation Theorem: BAR[BAR(A)] + BAR[BAR(B)] = BAR(A) + BAR(B)

Boolean Algebra and Logic Simplification

Q1: When are the inputs to a NAND gate, according to De Morgan's theorem, the output expression could be:

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Boolean Algebra and Logic Simplification

Q1: When are the inputs to a NAND gate, according to De Morgan's theorem, the output expression could be:

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