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Looking at the truth table, they're equivalent:

A   B      (A+B)  (A+A'B)
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1   1       1       1
1   0       1       1
0   1       1       1
0   0       0       0

But what manipulation can one do using basic identifies and laws to show that they're the same?

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    $\begingroup$ By A' you mean not A? I personally never used nor seen used that notation. I think $\overline{A}$ and $\lnot A$ are much more used around the world (or at least in the material I have seen). $\endgroup$ – Bakuriu Sep 18 '16 at 7:26
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    $\begingroup$ @Bakuriu it's a notation that's used more in computer science, out of the necessity of not being able to do overbar in ASCII. See: courses.cs.washington.edu/courses/cse370/99sp/lectures/02-Comb/… $\endgroup$ – C.J. Jackson Sep 18 '16 at 18:05
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Note with the laws of Boolean algebra, "addition" distributes over "multiplication" (just as multiplication would normally distribute over addition). Thus, we have $$ a + (a'\cdot b) = (a+a')\cdot (a+b) = 1(a+b) = a+b $$

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  • $\begingroup$ Thanks. It's easy to forget that multiplication distribute over addition in boolean algebra. $\endgroup$ – C.J. Jackson Sep 18 '16 at 5:19
  • $\begingroup$ You mean the other way. $\endgroup$ – Leif Willerts Sep 18 '16 at 9:27
  • $\begingroup$ Huh, I knew the law but it hadn't sunk in for me that it's really addition distributing over multiplication. +1 for phrasing it that way! $\endgroup$ – user541686 Sep 18 '16 at 9:55
  • $\begingroup$ @Leif what is it that you think I mean? $\endgroup$ – Ben Grossmann Sep 18 '16 at 14:23
  • $\begingroup$ I meant OP means addition distributes over multiplication - that that's the hard to forget part. $\endgroup$ – Leif Willerts Sep 19 '16 at 19:51
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$A+A'B=A(1+B)+A'B=A+AB+A'B=A+(A+A')B=A+B$

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