Recall that a ⇒ b ⇔ ¬a ∨ b. So we can write
q ⇒ (p ∧ r ) ⇔ ¬q ∨ (p ∧ r )
Then the negation of this would be, for instance,
¬(¬q ∨ (p ∧ r )) ⇔ ¬(¬q) ∧ ¬(p ∧ r )
… ⇔ q ∧ (¬p ∨ ¬r )
… ⇔ (¬p ∧ q) ∨ (q ∧ ¬r )
… ⇔ ¬(p ∨ ¬q) ∨ (q ∧ ¬r )
… ⇔ (p ∨ ¬q) ⇒ (q ∧ ¬r )
It's impossible to tell what kind of statement your program is expecting, but since there are 5 slots available, my money would be on q ∧ (¬p ∨ ¬r ), so long as ¬p and ¬r are options.
