Consider the following statement. ∀ integer d, if 6 d is an integer then d = 3. Which of the following is a negation for the statement? ∃ an integer d such that 6 d is an integer and d ≠ 3. ∃ an integer d such that 6 d is not an integer and d = 3. ∀ integer d, if 6 d is not an integer, then d ≠ 3. ∀ integer d, if d ≠ 3, then 6 d is not an integer. ∀ integer d, if d = 3, then 6 d is an integer.

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

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step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2

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step-by-step explanation: