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Can someone explain to me why in Boolean algebra

$$ f(x,y,z,t)=z+x'y+xy'+xt'+yt' =z+x'y+xy'+xt'$$

I have no clue why u can just leave out the last term, is it due to some absorption law.

Ps: $x'$ is the complement of $x$.

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  • $\begingroup$ The only interesting case is where $y$ is true and $t$ is false. In this case, either $x$ is true or $x$ is false. If $x$ is false, then $x'y$ is true. If $x$ is true, then $xt'$ is true. $\endgroup$ – Michael Burr Aug 16 '15 at 21:51
  • $\begingroup$ It's because $x'y+xt'+yt'=x'y+xt'.$ Try drawing a Venn diagram. $\endgroup$ – bof Aug 16 '15 at 21:54
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The consensus theorem helps reduce your expression. It is a very important method in digital design for reducing boolean expressions. I'll provide an algebraic proof below:

\begin{align} x^{\prime} y + x t^{\prime} + y t^{\prime} &= x^{\prime} y + x t^{\prime} + y t^{\prime} (x + x^{\prime}) \\ &= x^{\prime} y + x t^{\prime} + x y t^{\prime} + x^{\prime} y t^{\prime} \\ &= x^{\prime} y + x^{\prime} y t^{\prime} + x t^{\prime} + x y t^{\prime} \\ &= x^{\prime} y (1 + t^{\prime}) + x t^{\prime} (1 + y) \\ &= x^{\prime} y + x t^{\prime}. \end{align}

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For a Boolean algebra B let $B^*= \{ p^* |p \in B \}$ where $p^*= \{ q \in B |qp=q \}$ for $p \in B$. Then $(pq)^*=p^* \cap q^*$ and $(p+q)^*=p^* \cup q^*$ for $p,q \in B$,and $p'^*=B^* -p^* .$ So $B$ is isomorphic to the Boolean algebra $B^*$ ,where addition and multiplication in $B^*$ are $ \cup$ and $ \cap$, and $x'$ in $B^*$ means $B-x$ .So, if you consider $f$ as if plus and times are $\cup$ and $\cap$,and $p'$ is the complement of $p$, it works out easily .

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