For each of these compound propositions, use the conditional-disjunction equivalence (Example 3) to find an equivalent compound proposition that does not involve conditionals.
a) ¬p → ¬q
b) (p ∨ q) → ¬p
c) (p → ¬q) → (¬p → q)

See below

Step-by-step explanation:

The equivalence you are required to use is the following:

t→s ⇔ ¬t∨s

This also implies, negating both statements and using De Morgan's Laws

¬(t→s) ⇔ t∧¬s (*)

a) Here t=¬p and s=¬q. Hence ¬p → ¬q ⇔ ¬¬p∨¬q ⇔ p∨¬q

b) Here t=p∨q and s=¬p. Thus  (p ∨ q) → ¬p ⇔ ¬(p ∨ q) ∨ ¬p ⇔ (¬p∧¬q)∨¬p  (this last step is by De Morgan's Laws)

c) Here t= (p → ¬q) and s=(¬p → q), therefore

(p → ¬q) → (¬p → q) ⇔ ¬(p → ¬q)∨(¬p → q) ⇔ (p∧¬q)∨(¬¬p∨q) ⇔ (p∧¬q)∨(p∨q)  

We have to apply he equivalences twice. In the last part we used (*) to negate the conditional, and applied the equivalence to ¬p → q.

step-by-step explanation:

no solution.