I am having trouble converting this to its Disjunctive Normal Form.
(A ∨ ¬C ∨ D)∧ (B ∨ C ∨ ¬D) ∧(¬A ∨ B ∨ C)∧ (¬A ∨ B ∨ ¬D)
The 3 variables that exist in the brackets, is confusing me. How do the steps look like when I am trying to expand 3 variables with another 3.
Any help would be greatly appreciated.
Kind regards
Let me make the notation easier by using concatenation for "and", + for "or" and ' for complementation. So the expression is $$X=(a+c'+d)(b+c+d')(a'+b+c)(a'+b+d')$$ Use distributivity to obtain $$(a+c'+d)(b+c+d')=ab+ac+ad'+bc'+cc'+c'd'+bd+cd+dd'$$ and since $xx'=0$ it becomes $$ab+ac+ad'+bc'+bd+cd+c'd'$$ Also $$(a'+b+c)(a'+b+d')=a'+a'b+a'c+a'd'+b+bc+bd'+cd'$$ and by absorption ($x+xy=x$) it is just $a'+b+cd'$. So again by distributivity, $$X=(ab+ac+ad'+bc'+bd+cd+c'd')(a'+b+cd')$$ becomes $$aa'b+ab+abcd'+aa'c+abc+acd'+aa'd'+abd'+acd'+a'bc'+bc'+bcc'd'+a'bd+bd+bcdd'+a'cd+bcd+cdd'+a'c'd'+bc'd'+cc'd'$$ and simplifies to $$ab+abcd'+abc+abd'+acd'+a'bc'+bc'+a'bd+bd+a'cd+bcd+a'c'd'+bc'd'$$ and finally to $$ab+acd'+bc'+bd+a'cd+a'c'd'$$
Following the comment of other user using that $X+Y=X$ iff $XY=Y$, reduce to $$X=ab+acd'+bc'+a'cd+a'c'd'$$ since $$(ab+bc'+a'cd)bd=(a+c'+a'c)bd$$ and $$a+c'+a'c=a+(a+a')c'+a'c=(a+ac')+a'(c+c')=a+a'=1.$$