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I've been trying to interpret a logical proposition for several days. I need to simplify, I can't do it with the laws that I know, but using online tools I can find the result, but in the step by step a method appears that I don't know

The question is: In the next table what does it mean: "introduce Variables" (I know the other laws)

where does (A+A’) come from

the proposition is: [(A’B’) + (A’+C)’]’ (B’C’)’

The Solution is:

[(A’B’)+(A’+C)’]’ (B’C’)’ Defininition
(A+B)(A’+C’’)(B’C’)’ De morgan Law
(AA’ + AC + BA’ + BC) (B’C’)’ Distrib Law
(0 + AC + BA’ + BC) (B’C’)’ Complement Law
(AC + BA’ + BC) (B’C’)’ Identity Law
(AC + BA’ + BC) (B’’+C’’) De morgan Law
ACB + ACC +BA’B + BA’C + BCB + BCC Distrib Law
ACB + CA + BA’B + BA’C + BCB + BCC Idemp Law
AC + BA’B + BA’C + BCB + BCC Absorpt Law
AC + BA’ + BA’C + BCB + BCC Idemp Law
BA’ + AC + BCB + BCC Absorpt Law
BC + BA’ + AC Idemp Law
BC (A+A’) + BA’ + AC Introduce Variables
BCA + BCA’ + BA’ + AC Distrib Law
AC + BA’ Absorpt Law
AC + BA’ Answer

Note: Other sites call this step: Consensus

Thanks a lot, my head is going to explode, I can't find anything on the internet to guide me.

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  • $\begingroup$ I suppose you know that $A+A'=1$, and that $X1=X$, for all $X$. So $BC=BC(A+A')$. $\endgroup$
    – amrsa
    Commented Feb 14, 2021 at 10:10
  • $\begingroup$ Hi @amrsa, first of all thanks for your quick response, i understand what you mention, but why does it apply only to BC and not to the others? is it for convenience? I have to explain the step by step of why things happen, and I don't know how to explain it, starting because I don't even know what to call it, can i justify this step as "algebraic manipulation"??. Tanks for the help and patience $\endgroup$
    – Pilaf
    Commented Feb 14, 2021 at 18:48
  • $\begingroup$ What is called the Consensus theorem is not that identity alone, but $$BC+BA'+AC=AC+BA'.$$ What you have in those steps is its proof (for this particular case). That was the chosen step because it was convenient, yes! So it's a form of algebraic manipulation but it was done with some insight. Probably whoever did it knew about the theorem. $\endgroup$
    – amrsa
    Commented Feb 14, 2021 at 20:10
  • $\begingroup$ @amrsa, thanks a lot, that's the answer that i needed, I can manipulate the equation to convenience as long as it does not affect its values $\endgroup$
    – Pilaf
    Commented Feb 14, 2021 at 20:36

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