0
$\begingroup$

I learning how to convert to disjunctive normal forms, I have the following,

(¬(p → q) → (q ∧ ¬r))

I understand any p→q can be represented as (¬p)∨q, therefore if I image the above as just that I can break each section down, resulting in:

1. p → q == (¬p)∨ q, therefore (¬(p → q) == ¬((¬p) ∨ q
2. ¬((¬p) ∨ q → (q ∧ ¬r)) == ¬(¬((¬p)∨q)) ∨ (q ∧ ¬r )

Therefore the disjunctive form is:
¬(¬((¬p)∨q)) ∨ (q ∧ ¬r )

The truth table for this seems to match, is this correct? I have a feeling I have left a step out, maybe repetition of the NOT could be fixed?

$\endgroup$
  • $\begingroup$ @RodrigodeAzevedo from my textbook "Any logical expression can be written as the sequence of or's of its min-terms. A min-term is an and of all the input variables. An expression written in this form is said to be in disjunctive normal form." $\endgroup$ – Ari Victor May 12 at 8:09
  • 2
    $\begingroup$ Yes, the repetition of two NOTs can be cancelled : $\lnot (\lnot \varphi) \equiv \varphi$. $\endgroup$ – Mauro ALLEGRANZA May 12 at 8:11
1
$\begingroup$

$\neg(p\Rightarrow q) \Leftrightarrow \neg(\neg p\vee q)$

and so

$\neg(p\Rightarrow q) \Rightarrow (q\wedge \neg r)$

is equivalent to

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

is equivalent to

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

is equivalent to

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

By absorption $a\vee (a\wedge b) \Leftrightarrow a$, it is equivalent to

$\neg p\vee q$.

$\endgroup$
1
$\begingroup$

Using SymPy:

>>> from sympy import * 
>>> p, q, r = symbols('p q r')
>>> phi = Not(p >> q) >> (q & Not(r))

Converting to DNF and simplifying:

>>> to_dnf(phi, simplify=False)
q | ~p | (q & ~r)
>>> to_dnf(phi, simplify=True)
q | ~p

Hence, we have the DNF formulas

$$q \vee \left(q \wedge \neg r\right) \vee \neg p \equiv q \vee \neg p$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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