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The boolean logic equation for a 2-to-1 multiplexer is \$(A.\bar{S})+(B.S)\$, where \$A\$ is the first input and \$B\$ is the second input. Is this logic equation equivalent to \$(A.S)+(B.\bar{S})\$, given that the inputs \$A\$ and \$B\$ have the same order? Thanks in advance!

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2 Answers 2

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The logic is the same, in that both formulae describe a 2-to-1 multiplexer that will select either \$A\$ or \$B\$ depending on the state of \$S\$. However, the two forms are different in that \$(A\cdot\bar{S})+(B\cdot S)\$ will select \$B\$ when \$S\$ is true, but \$(A\cdot S)+(B\cdot \bar{S})\$ will select \$A\$ when \$S\$ is true.

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If you set A to 0 and B to 1 and S to 0 you get a value of 0 for the output. If you invert S you effectively swap A and B and get 1 at the output. So this is not equivalent.

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