# Are these logical predicate translations same, thus correct?

In this problem in Problem set(1) P(1.c) of MIT's 6.042:

Translate the following sentences from English to predicate logic.

The domain that you are working over is X, the set of people. You may use the functions S(x), meaning that “x has been a student of 6.042,” A(x), meaning that “x has gotten an ‘A’ in 6.042,” T(x), meaning that “x is a TA of 6.042,” and E(x, y) , meaning that “x and y are the same person.”

(c) [6 pts] There are no people who are 6.042 TA’s who did not get A’s in 6.042.

¬∃x ∈ X : T(x) ∧ (¬A(x))

but, I came up with the following translation which i feel is the same logically. is it so? if not, how?

∀x ∈ X : T(x) ∧ A(x)

• No. Yours says “every person is a TA of 6.042 and has gotten an A in 6.042.” There’s doesn’t say anything about whether or not any person is a TA of 6.042; it just says that anyone who is a TA (if there are any) has gotten an A in 6.042.
– Joe
Commented Apr 17, 2020 at 15:31
• No it is not. If you move inside the negation sign, from ¬∃x you get ∀x¬. Now, ¬ (T(x) ∧ (¬A(x))) is T(x) → A(x) Commented Apr 17, 2020 at 15:57

$$$$\tag{1} \lnot\exists x\in X:T(x)\land(\lnot A(x))$$$$ $$$$\tag{2} \forall x\in X:T(x)\land A(x)$$$$
Consider the case where T(x) is true, and A(x) is false. In this case, $$(1)$$ is true, however $$(2)$$ is false, by virtue of the logical AND in each of them.
In general, $$\big(\lnot\exists x\in X:P(x)\big)\equiv\big(\forall x\in X:\lnot P(x)\big)$$. So an answer equivalent to $$(1)$$ that uses a universal quantifier, as opposed to an existential one, is $$(3)$$, which is equivalent to $$(4)$$.
$$$$\tag{3} \forall x\in X: \lnot T(x)\lor A(x)$$$$ $$$$\tag{4} \forall x\in X: T(x)\implies A(x)$$$$