A and ( A or C ) = A


A or A and C = A

How do these identities work?

Using the rule

A and ( B or C ) = A and B or A and C

For the first identity, I get

A and A or A and C = A or A and C

How is the C eliminated?

Mental substitution shows that it will equal A, but can it be algebraically shown.

Equation 2, too.

  • 5
    $\begingroup$ Why Don't you just make a truth table? $\endgroup$
    – joanolo
    Commented Feb 27, 2017 at 0:18
  • $\begingroup$ Because, as I wrote in the question, I've already done it, mentally, and noticed that it works. I want to know why, and how it can be proven (without brute force). $\endgroup$
    – Tobi
    Commented Feb 27, 2017 at 0:23
  • 4
    $\begingroup$ What do you mean by "how to they work"? They don't "work"; they just are. What do you mean by "A or A and C"? Don't rely on everybody knowing what operator precedence you're using. $\endgroup$ Commented Feb 27, 2017 at 0:24
  • $\begingroup$ AND, before, OR, David. $\endgroup$
    – Tobi
    Commented Feb 27, 2017 at 0:26
  • 1
    $\begingroup$ @Tobi, an extra pair of parentheses for explicit readability killed exactly no one. :) $\endgroup$ Commented Mar 1, 2017 at 6:56

2 Answers 2


Here is one way to prove the first identity: $$ A \land (A \lor C) = (A \lor 0) \land (A \lor C) = A \lor (0 \land C) = A \lor 0 = A. $$ The second identity has a similar proof. Alternatively, you could use duality to deduce it from the first identity.


Here's one way of thinking how these identities "work". Of the first one, when A is false, A and anything is false; when A is true, A or C is true, and the whole thing is true too; therefore being equal A in both cases. Similarly, the second, when A is true, then A or anything is true; when A is false, A and C is false, and the whole expression is also false.


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