Disjunctive normal form of (¬(p → q) → (q ∧ ¬r)) 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?

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

and so

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

is equivalent to

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

is equivalent to

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

is equivalent to

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

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

$\
eg p\vee q$.