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I'm getting stuck at the following boolean expression.

$$z + (x'y) + (xy') + (xt') + (yt')$$

In my solutions it's simplified and the $(yt')$ term is gone. How do they simplify this? I really cant see it....

Many thanks!

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  • $\begingroup$ as i see it you're only allowing x and y to be different from the 2nd and 3rd terms, so only one of the 4th and 5th can be the case $\endgroup$ – danimal Jun 10 '15 at 17:11
  • $\begingroup$ I think i understand what you mean but is there a simple theorem or rule to simplify it? $\endgroup$ – Stekke Jun 10 '15 at 17:18
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This comes from the Consensus Theorem. Notice that: \begin{align*} x'y + xt' + yt' &= x'y + xt' + (1)yt' \\ &= x'y + xt' + (x + x')yt' \\ &= x'y + xt' + (xyt' + x'yt') \\ &= (x'y + x'yt') + (xt' + xyt') \\ &= x'y(1 + t') + xt'(1 + y) \\ &= x'y(1) + xt'(1) \\ &= x'y + xt' \\ \end{align*}

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  • $\begingroup$ Thanks a lot! This is what i needed! $\endgroup$ – Stekke Jun 10 '15 at 17:26

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