0
$\begingroup$

Like many before me I am trying to write a proposition in DNF. I have tried to follow all of the previous articles but I'm unsure who to correctly transfer from CNF to DNF. I have done the following

Prop - (p→q)∧(q→p)

= (p→q) = (p∧q)∨(⌐p∧q)∨(⌐p∧⌐q) = (p∧q)∨⌐p = ⌐p∨q

= (q→p) = (q∧p)∨(⌐q∧p)∨(⌐q∧⌐p) = (q∧p)∨⌐q = ⌐q∨p

= (⌐p∨q)∧(⌐q∨p)

From the above I have a conjunction that needs to be changed, do I simply work out the values of each prop and repeat the process for the disjunction?

  • 0
    [Here](https://www.cs.jhu.edu/~jason/tutorials/convert-to-CNF.html) is a well-written lecture note you might want to go through. If you "multiply out" your last expression, you directly get the DNF "!p!q v pq". A quick alternative way would be a truth-table plus Karnaugh map. – 2017-01-10
  • 0
    @AxelKemper thanks for responding. Could this be simplified further or do I stop? Do they not cancel each other out? – 2017-01-10

1 Answers 1

0

Once you get to $(\neg p \lor q) \land (\neg q \lor p)$ you can proceed by distributing $\land$ over $\lor$:

$(\neg p \lor q) \land (\neg q \lor p) \Leftrightarrow$ (Distribution)

$((\neg p \lor q) \land \neg q) \lor ((\neg p \lor q) \land p)) \Leftrightarrow$ (Distribution * 2)

$(\neg p \land \neg q) \lor (q \land \neg q) \lor (\neg p \land p) \lor (q \land p) \Leftrightarrow$ (Complement * 2)

$(\neg p \land \neg q) \lor \bot \lor \bot \lor (q \land p) \Leftrightarrow$ (Identity * 2)

$(\neg p \land \neg q) \lor (q \land p)$

This end result makes sense: $p \leftrightarrow q$ is true if and only if $p$ and $q$ have the same truth value, i.e. either they are both true, or they are both false, i.e. $(q \land p) \lor (\neg p \land \neg q)$. In fact, this is how I usually rewrite a biconditional in 1 step. It's a good equivalence to put in your 'logic tool box'!