I have a Boolean algebra with some elements $a,b,c$. I have to show this:

$(a ∧ b) ∨ (a′ ∧ c) ∨ (b ∧ c) = (a ∧ b) ∨ (a′ ∧ c)$.

Now I have done other such proofs before but this one I got lost in. I see Boolean algebras have a cancellation law, so my guess was I had to cancel the terms that are on both sides, to get:

$b ∧ c = O_B.$

Since this ended up looking more like an equation than an identity - I could use some help. I would especially like to know why the cancellation method fails. Thanks!

  • 1
    $\begingroup$ Hint: $b\land c \rightarrow (a \land b)\lor(a' \land c)$ $\endgroup$ – SBareS Jan 4 '15 at 22:03
  • $\begingroup$ How is this derived? $\endgroup$ – Snowflake Jan 4 '15 at 23:42
  • $\begingroup$ Well, if $bc=1$ (meaning that both $b$ and $c$ are equal to $1$), then $ab+\overline{a}c$ will be $1$ anyhow. $\endgroup$ – Shahar Jan 5 '15 at 1:40
  • $\begingroup$ I'm sorry but I don't get the notation... but I believe you just repeated what @SBareS said, I wanted to know what rules are used to derive that since I wouldn't know how to use the distributive law on that. I get it on an intuitive level. $\endgroup$ – Snowflake Jan 5 '15 at 11:35
  • $\begingroup$ @Snowflake You can set $b=c=1$ and then $(a\land 1)\lor(a'\land 1)=a\lor a'= 1$. $\endgroup$ – SBareS Jan 5 '15 at 13:10

Here's a hint: In a Boolean algebra, $a \vee a = a \vee a \vee a$ for all elements $a$, but that doesn't mean that $O_B = a$ for all $a$. Thus I think you should look back at what kind of cancellation laws are valid.


  • $\begingroup$ What it says in my book is that $a V b = a V c → b = c$, and same thing with ∧. Both this and your example make intuitive sense to me, but since it's hard when the formulas are more complicated I'd need to know what's the substantial difference. $\endgroup$ – Snowflake Jan 4 '15 at 23:45

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