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.
Answer provided is this:
¬∃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)
You can show that the answers (1) and (2) are not equivalent by finding a truth assignment of T(x) and A(x), where they disagree.
\begin{equation}\tag{1} \lnot\exists x\in X:T(x)\land(\lnot A(x)) \end{equation} \begin{equation}\tag{2} \forall x\in X:T(x)\land A(x) \end{equation}
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 $(4)$.
\begin{equation}\tag{3} \forall x\in X: \lnot T(x)\lor A(x) \end{equation} \begin{equation}\tag{4} \forall x\in X: T(x)\implies A(x) \end{equation}