Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was working on an examples from my textbook concerning transforming formulae into disjunctive-normal form (DNF) until I found an expression that I cannot solve. I hope somebody can help me transform the following statement into DNF:

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

share|cite|improve this question
up vote 4 down vote accepted

$$ (\lnot q \lor r) \land ( q \lor \lnot r)\tag{1}$$ $$[(\lnot q \lor r) \land q] \lor [(\lnot q \lor r) \land \lnot r]\tag{2}$$ $$[(\lnot q \land q) \lor (r \land q)] \lor [(\lnot q \land \lnot r) \lor (r \land \lnot r)]\tag{3}$$ $$ \text{False} \lor (r \land q) \lor (\lnot q \land \lnot r) \lor \text{False}\tag{4}$$

$$(q \land r) \lor (\lnot q \land \lnot r)\tag{5}$$

Note that $(5)$ is is in disjunctive normal form (DNF), and is equivalent to $(1)$.

$(1) \to (2)$: Distribution;
$(2) \to (3)$: Distribution, twice;
$(3) \to (4)$: Contradictions $\lnot q \land q \rightarrow \text{False}$ and $\lnot r \land r\rightarrow \text{False}$;
$(4) \to (5)$: Simplification (removal of contradictory disjuncts).

Observation: both $(1)$ and $(5)$ are equivalent to $q \leftrightarrow r$:
To see this, you can rewrite $(1)$ as: $$(\lnot q \lor r) \land (\lnot r \lor q) \iff (q \rightarrow r) \land (r\rightarrow q) \iff (q \leftrightarrow r).$$

$$q \leftrightarrow r\; \text{ is true if and only if }\;(q \land r) \text{ is true or}\; (\lnot q \land \lnot r)\text{ is true.}$$ $$\text{That is,}\;q \leftrightarrow r\; \text{ is true if and only if }\;\;(q \land r) \lor (\lnot q \land \lnot r).\tag{5} $$

p | q |$\;\lnot q \lor r$ | $q \lor \lnot r\;$|$\;(\lnot q \lor r) \land (q\lor \lnot r)$
T | T | $\quad$T $\quad$|$\quad$ T $\quad$| $\quad\quad\quad\quad$T$\quad\quad\leftarrow\quad\;\; (q \land r)$
T | F | $\quad$F $\quad$|$\quad$ T $\quad$| $\quad\quad\quad\quad$F
F | T | $\quad$T $\quad$|$\quad$ F $\quad$| $\quad\quad\quad\quad$F
F | F | $\quad$T $\quad$|$\quad$ T $\quad$| $\quad\quad\quad\quad$T$\quad\quad\leftarrow\quad (\lnot q \land \lnot r)$

share|cite|improve this answer

Make a table

 q  r  result
 0  0     1
 0  1     0
 1  0     0
 1  1     1

Now you can get the DNF formula, by "or-ing" all the rows with result=1: $$ (\lnot q \land \lnot r ) \lor ( q \land r )$$

share|cite|improve this answer

Think of it as an expression with $+$ and $\cdot$ and use the distributive law (I'll use $q'$ as notation for $\neg q$:

$(q' + r)(q + r') = q'q + q'r' + rq + rr' = q'r' + rq$

I cancelled $q'q$ and $r'r$ because $q$ cannot be true and false (and same for $r$). With your symbols, this gives

$(\neg q \wedge \neg r) \vee (r \wedge q)$

You can (and should) check that this is correct with a truth table.

share|cite|improve this answer

By the distibution law one obtains $(\lnot q \lor r) \land ( q \lor \lnot r) \leftrightarrow ((\lnot q \lor r) \land q) \lor ((\lnot q \lor r) \land \lnot r) \leftrightarrow (\lnot q \land q) \lor (r \land q)\lor (\lnot q \land \lnot r) \lor (r \land \lnot r)\leftrightarrow (r \land q)\lor (\lnot q \land \lnot r)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.