Answer:
The correct option is the first:
∃ an integer d such that 6d is an integer and d ≠ 3
Step-by-step explanation:
If 6d is an integer, then d = 3
Let's divide the statement into two:
Let "If 6d is an integer" be p and "then d = 3" be q.
The statement is a conditional statement. Therefore, we have
p ⇒ q
The negation of p ⇒q, ¬(p ⇒ q) ≡ p ∧ ¬q
I have attached an image proving the above.
Using this: ¬(p ⇒ q) ≡ p ∧ ¬q
Then, we leave p as it is and negate q.
P is "if 6d is an integer"
q is "then d = 3," hence ¬q is "then d ≠ 3."
Therefore, the negation of "If 6d is an integer, then d = 3" is "If 6d is an integer, then d ≠ 3."
The correct option is the first:
∃ an integer d such that 6d is an integer and d ≠ 3
