4
$\begingroup$

In this problem in Problem set(1) 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.”

  • (a) [6 pts] There are people who have taken 6.042 and have gotten A’s in 6.042
  • (b) [6 pts] All people who are 6.042 TA’s and have taken 6.042 got A’s in 6.042
  • (c) [6 pts] There are no people who are 6.042 TA’s who did not get A’s in 6.042.
  • (d) [6 pts] There are at least three people who are TA’s in 6.042 and have not taken 6.042

This is my solution: $$ \exists x \in X. A(x) $$ $$ \forall x \in X. T(x) \rightarrow A(x) $$ $$ \forall x \in X. \neg (T(x) \wedge \neg A(x) ) $$ $$ \exists x \in X. ( T(x) \wedge \neg S(x) ) $$

Are these correct? And if so how could I've used E(x,y)??

$\endgroup$
0

2 Answers 2

2
$\begingroup$

It is a matter of what is used in your course, but because the domain is the set of all people, I would leave out the $\in X$ everywhere.

For (b), it should be something like $\forall x((T(x)\land S(x))\rightarrow A(x))$.

For (d), the following will work: $$\exists x\exists y\exists z(T(x)\land T(y)\land T(z)\land \lnot E(x,y)\land \lnot E(y,z)\land \lnot E(z,x) \land \lnot S(x)\land \lnot S(y)\land \lnot S(z)).$$ (Formally, this is not quite right, since we need many more parentheses.)

$\endgroup$
2
$\begingroup$

I know this has an accepted answer. Just to add for the sake of being helpful. If you look here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/MIT6_042JF10_chap01.pdf

In the readings there is a section on predicates. The explanation I found most helpful is on page 14 & 15. Every American has a dream was represented as H(a,d) where H was to shown to define the predicate that the American "has" the dream. I think E(x,y) is used similarly here. where you define the terms: 1. There exists at least 1 person who has taken 6.042.
2. There exists at least 1 person who has gotten an A.
3. For E(x,y) There is a case that satisfies the predicate that the person who got the A and the person who took 6.042 are the same person. In that example it was written after the exists statements.

Hope that helps.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .