Answer
See step by step work for solution
Work Step by Step
We start with $¬p→(q→r)$
Use Logical Equivalence twice: (p→q)≡(¬p∨q)
$$¬p→(q→r)≡¬(¬p)∨(q→r)≡¬(¬p)∨(¬q∨r)$$Use Double Negation:
$$≡p∨(¬q∨r)$$
Use Associative Law:$$(p∨¬q)∨r)$$
Use Commutative Law:$$(¬q∨p)∨r)$$
Use Associative Law:$$¬q∨(p∨r)$$
Use Logical Equivalence:(p→q)≡(¬p∨q)
$$≡q\rightarrow (p∨r)$$
We have thus derived that $¬p→(q→r)$≡$q\rightarrow (p∨r)$.
