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so I'm stuck on what seems like the very last steps of the problem. I'll leave out the whole problem and just start from where I got stuck at in the very end.

How can I show that c'd' + c is equal to c + d' using the identities? It's probably just flying over my head...

user215717
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1 Answers1

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Hint: $\; c'd' + c \; = \; (c'+c)(d'+c) \;$ by Distribution.

Can you see what you do next?

Graham Kemp
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