Answer:
(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
x + 2