DNF conversion of this statement

Question

How to convert this to DNF?

$(x\lor{\neg{y}\lor{z}})\land{(\neg{x}\lor{\neg{z}})}$

I have tried de morgans and have got no where. I'm pretty sure its the distributive law but cant work out the steps to get in the form to do that.

Answer 1

I never heard of DNF before, looked it up here. I take it you just want the simplest boolean out of this.

This is a case that works like distributive multiplication of two sums.

$$(x\lor{\neg{y}\lor{z}})\land{(\neg{x}\lor{\neg{z}})}\\ =x(\neg x\lor \neg z)\lor \neg y(\neg x\lor \neg z)\lor z(\neg x\lor \neg z)\\ =(x\neg x \lor x\neg z)\lor (\neg y\neg x \lor \neg y\neg z)\lor z\neg x \lor z\neg z)\\ $$ If there were no mistakes in this expansion, the Veitch diagram below helps with the solution. For shorthand, a slightly different notation is used. Note that hte conditions $x\neg x$ and $z\neg z$ cannot exist so they are to be ignored.