I was able to find a demonstration,
starting from the right part,
I wanted to start from the left part, in order to check if I understood well my knowdlege,
Starting from left or right does make any difference in how it's complex the resolution? And, in case yes, what criteria should I use to identify which part is simpler?
My current steps are these:
(intuition, with the expanded form, I can have some chance to re-group as I need, I will add the missing term but with neutral values)
ab+bc+ca = ab(c+'c) + bc (a+'a) + ca (b+'b)
................ = abc + ab'c + abc + 'abc + abc + a'bc
................ = how should I proceed?
(the single quote is not)
The steps from the right are these:
(a+b)(b+c)(c+a) = a(b+c)(c+a)+b(b+c)(c+a)
........................... = abc+aab+acc+aac+bbc+abb+bcc+abc
........................... = abc+ab+ac+ac+bc+ab+bc
........................... = abc+ab+ac+bc
........................... = ab(c+1) + ac + bc
........................... = ab+ac+bc
Following the suggestion in the comment, i.e. how's supposed to deduct that starting from ab+ac+bc, I should add ab(c+1)? The general intuition here is what's missing, thank you
Distribution works both ways, also over AND. To mimic your attempt from right to left:
$$\begin{align*} ab+bc+ca &= (a+bc+ca)(b+bc+ca)\\ &= (a+b+c)(a+b+a)(a+c+c)(a+c+a)(b+b+c)(b+b+a)(b+c+c)(b+c+a)\\ &= (a+b+c)(a+b)(a+c)(a+c)(b+c)(a+b)(b+c)\\ &= (a+b+c)(a+b)(a+c)(b+c)\\ &= (a+b+c0)(a+c)(b+c)\\ &= (a+b)(a+c)(b+c) \end{align*}$$