I don't really get the absorption law, specifically in this case:

$$ (\lnot p \lor q) \land (\lnot r \lor q) \equiv (\lnot p \land \lnot r) \lor (\lnot p \land q) \lor (q\land \lnot r) \lor (q \land q) $$

The absorption law is used here: $ (\lnot p \land q) \lor (q\land \lnot r) $

I don't really understand the use of absorption law in this case, and if someone could explain it to me on a similar example or something, that would be highly appreciated.

  • $\begingroup$ What do you mean, "the absorption law is used here"? Do you mean that it is used to change $ (\lnot p \land \lnot r) \lor (\lnot p \land q) \lor (q\land \lnot r) \lor (q \land q) $ into $ (\lnot p \land q) \lor (q\land \lnot r) $ ? Or do you mean something else? $\endgroup$ – MJD Nov 1 '14 at 15:52
  • $\begingroup$ I mean that it is used here: $ (\lnot p \land q) \lor (q\land \lnot r) $ When we use the absorption law, it cancels out, but I don't get it how? So in the end we only get: $ \lnot p \land \lnot r \lor q $ $\endgroup$ – mythic Nov 1 '14 at 17:54

With :


you first apply $(q \land q) \equiv q$ to simplify it :


Consider now : $(q∧¬r)∨q$; applying Absorption Laws : $p∨(p∧q) \equiv p$, we have that it is equivalent to $q$.

Thus :

$[(¬p∧¬r)∨(¬p∧q)]∨[(q∧¬r)∨q] \equiv (¬p∧¬r)∨(¬p∧q)∨q$.

Apply it again to : $(¬p∧q)∨q$, which is again equivalent to $q$, to get :

$(¬p∧¬r)∨[(¬p∧q)∨q] \equiv (¬p∧¬r)∨q$.

Regarding absorption :

$p∨(p∧q) \equiv p$

you can prove it with Natural Deduction or verify it via truth table :

  • if $p$ is TRUE, then $p \lor (p \land q)$ is TRUE, by truth table for $\lor$;

  • if $p$ is FALSE, then $(p \land q)$ is FALSE, by truth table for $\land$; so both $p$ and $(p \land q)$ are FALSE, and thus $p∨(p∧q)$ is FALSE, by truth table for $\lor$.

  • $\begingroup$ Wow, that clears things up. We were doing this example in college and for some reason we always skip a few steps. I thought we somehow only used the absorption law once, so I really didn't get it. Thank you very much for the explanation, I really appreciate it. $\endgroup$ – mythic Nov 1 '14 at 19:08

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