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I have to find out the DNF of $$f(x,y,z)=x\wedge\left(\overline y\vee\overline z\right),$$ being $f$ a function of Boole.

$$\begin{array}{ll} f(x,y,z)&=x\wedge \left(\overline y\vee\overline z\right)\\ &=\left(x\wedge\overline y\right)\vee\left(x\wedge\overline z\right)\\ &=\left(x\wedge\overline y\wedge 1_A\right)\vee\left(x\wedge\overline z\wedge 1_A\right)\\ &=\left(x\wedge\overline y\wedge \left(z\vee\overline z\right)\right)\vee\left(x\wedge\overline z\wedge\left(y\vee\overline y\right)\right)\\ &=\left(x\wedge\overline y\wedge z\right)\vee\left(x\wedge\overline y\wedge\overline z\right)\vee\left(x\wedge\overline z\wedge y\right)\vee\left(x\wedge\overline z\wedge\overline y\right)\\ &=\left(x\wedge\overline y\wedge z\right)\vee\left(x\wedge\overline y\wedge\overline z\right)\vee\left(x\wedge\overline z\wedge y\right), \end{array}$$

being $1_A$ the last element.

Is it right? Thanks!

manooooh
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    That looks fine .. though $\left(x\wedge\overline y\right)\vee\left(x\wedge\overline z\right)$ is already in DNF. But if you need to make sure that all your terms refer to all variables involved, then you're doing exactly the right thing – Bram28 Feb 22 '18 at 18:01
  • @Bram28 Thank you! I thought that always in each term should appear all the variables of the function, I will consider that :). – manooooh Feb 22 '18 at 18:06
  • @Maooooh No, no need for that. In fact, if the whole expression would simplify to just $x$, it would be in DNF. – Bram28 Feb 22 '18 at 18:42

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