Boolean algebra expression simplification in conjuction normal form

Question

Is there a way to simplify this expression:

(!a || !c || b) && (!a || b) && (a) && (!a || !b || c) && (!b || !d || a) && (d || !c || !b) && (!d || e)

This calculator returned me expression:

(a) && (b) && (c) && (d) && (e)

I woud like to ask you, if anybody know how to simplify it with steps.

Thank you :)

Have nice day!

Answer 1

The main sources of simplification here are:

• $(\lnot x\lor y) \land x=(\lnot x\land x)\lor (y\land x)=0\lor(x\land y)=x\land y$

• $(\lnot x\lor\lnot y\lor z) \land x\land y=x\land(\lnot y\lor z)\land y=x\land y\land z$

And also commutativity and associativity of $\lor$ and $\land$.

Let's split your expression into pieces $A\land B\land C\land D\land E\land F\land G$

$\begin{array}{l} A=(\lnot a\lor\lnot c\lor b)\\ B=(\lnot a\lor b)\\ C=a\\ D=(\lnot a\lor\lnot b\lor c)\\ E=(\lnot b\lor\lnot d\lor a)\\ F=(\lnot b\lor\lnot c\lor d)\\ G=(\lnot d\lor e)\\ \end{array}$

So now this is a chain reaction

$B\land C=a\land b\\ B\land C\land D=a\land b\land c\\ B\land C\land D\land A=a\land b\land c\\ B\land C\land D\land A\land F=a\land b\land c\land d\\ B\land C\land D\land A\land F\land E=a\land b\land c\land d\\ B\land C\land D\land A\land F\land E\land G=a\land b\land c\land d\land e\\ $