boolean algebra reduction question hi im having a lot of trouble proving this boolean expression. Im getting many differing answers so I assume I must be going about it in the wrong way.
To explain, I'm trying to negate the whole LHS to split up the brackets and then trying to use the rules to reduce the expression from there. I must be using de morgans wrong or something since my answers vary. Can anyone give a hand.
$$((A\land B)\lor C\lor(\lnot D\land E))\land((A\land D)\lor\lnot C\lor (D\land E))\land((A\land B)\lor(A\land D)\lor E)\equiv((A\land B)\lor C\lor(\lnot D\land E))\land((A\land D)\lor\lnot C\lor(D\land E))$$
 A: You do not need deMorgan's.
This is an example of consensus. $(X + Y) • (\overline X + Z) • (Y + Z) = (X + Y) • (\overline X + Z)$
Take the right side of the equation.
$$\color {red}{(}(A\land B)\lor C\lor(\lnot D\land E)\color {red}{)}\land\color {red}{(}(A\land D)\lor\lnot C\lor (D\land E)\color {red}{)}$$
$${(}AB+ C+\overline D E)\ (AD+ \overline C + DE)$$
Notice $C$ and $\overline C$.  Apply consensus.  Remove $C$ and $\overline C$ and simplify.
$$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + \overline D E + AD  + DE )$$
$$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + AD + (\overline D + D)E )$$
$$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + AD + E)$$
$$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + AD + E) = (AB+ C+\overline D E)\ (AD+ \overline C + DE )$$
Laws and Theorems of Boolean Algebra
A: $((A\land B)\lor C\lor(\lnot D\land E))\quad =\quad P$
$((A\land D)\lor\lnot C\lor (D\land E))\quad =\quad Q$
$((A\land B)\lor(A\land D)\lor E)\,\;\;\quad =\quad R$    
$(P\land Q\land R)\equiv(P\land Q)$    
$(P\land Q\land R)\equiv(P\land Q\land 1[true])$    
The only way the whole shebang is true is if $R\quad$ END OF SENTENCE 
