Problem: Any professor who is bored by everyone, bores all/her students.
Domain: Set of all people
Predicates
- $P(x): x$ is a professor
- $B(x,y): x$ bores $y$
- $S(x,y): x$ is a student of $y$
Answer: $\forall x[P(x) \wedge \forall zB(z,x) \to \forall y(S(y,x) \to B(x,y))]$
I can understand the first conjunction part but it is hard to understand why there is implication between S(y,x) → B(x,y).
My original answer: $\forall x[P(x) \wedge \forall zB(z,x) \to \forall y(S(y,x) \wedge B(x,y))]$
Why is the implication correct, while conjunction is wrong ?
Any help is appreciated. Thank you.
The difference between
$$\forall x (P(x) \rightarrow Q(x))$$
and
$$\forall x (P(x) \land Q(x))$$
is tripping up a lot of beginning students of logic.
The first statement says "All P's are Q's"
The second statement says "Everything is both a P and a Q"
The first statement occurs more often in practice. For example, "All cats are furry", "All prime numbers greater than two are odd", and "All beginning students of logic struggle with the quantifiers" all follow the first pattern, where you use a $\rightarrow$. Indeed, if we were to use the $\land$ instead, we would get: "everything is a cat and furry", "Every number is both prime and odd", and "Everyone is a beginning student of logic and struggles with quantifiers"... which are clearly very unusual sentences.
And so it is with your sentence: if you use the $\land$, you end up saying that everything or everyone (including the professor themselves!) is both a student of the professor and is being bored by that professor ... which is clearly not what you want. What you want, is that all students of the professor are being bored by that professor, which follows the conditional form: If someone is a student of the professor, then that person will be bored by the professor.