a. Show that the following statement forms are all logically equivalent. p → q ∨ r, p ∧ ∼q → r, and p ∧ ∼r → q b. Use the logical equivalences established in part (a) to rewrite the following sentence in two different ways. (Assume that n represents a fixed integer.) If n is prime, then n is odd or n is 2.

(a) if n is prime, then n is odd or n is 2

(b) if n is prime and n is not odd, then n is 2

(c) if n is prime and n is not 2, then n is odd

Step-by-step explanation:

a) p → q ∨ r

b) p ∧ ∼q → r

c) p ∧ ∼r → q

Lets show that (a) implies (b) and (c). (a) says that if property p is true, then either q or r is true, thus, if p is true we have:

If the condition of (b) applies (thus q is not true), we need r to be true because either q or r were true because we are assuming (a) and p. Hence (b) is true If the condition of (c) applies (r is not true), since either r or q were true due to what (a) says, then q neccesarily is true, hence (c) is also true.

Now, lets prove that (b) implies (a)

If p is true and property (b) is true, then if q is true, then either q or r are true thus (a) is correct. If q is not true, then property (b) claims that, since p is true and q not, r has to be true, therefore (a) is valid in this case as well, hence (a) is also true.

(c) implies (a) can be proven with similar argument, changing (b) for (c), q for r and r for q.

With this we prove that the 3 properties are equivalent.

For the rest of the exercise, we have

property p: n is prime property q: n is odd property r: n is 2

Translating this, we obtain (a), (b) and (c)

(a) if n is prime, then n is odd or n is 2

(b) if n is prime and n is not odd, then n is 2

(c) if n is prime and n is not 2, then n is odd