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So I have this FOL question with the following key:

A-forall

E-there exist

|-or

&-and

B(x):x is a basketballer

F(x):x is a footballer

S(x): x is a swimmer

H(x,y):x is more lazy than y

Now, we were asked to represent "Only basketballers and footballers are more lazy than every swimmer". I interpreted it as an Only A's are B's situation and according to what I know, you represent that by Ax(B(x)->A(x)).

So that is what I did. Ax( (B(x)|F(x)) -> Ay(S(y) -> H(x,y)) ). The system said it is wrong. I switched it around to 'if you are an swimmer and there exists someone more lazy than you, then that person is either a footballer or a basketballer'. Still wrong

It has been downhill form there. I have checked every other way I can think about this. Nothing is working. I thought the every was the issue but I have checked like three textbooks and none of them that have a part on symbolizing "every" has done anything other than a forall.

Is there something I am missing?

1 Answer 1

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Your initial intuition is correct that "only A's are B's" is represented as Ax(B(x) -> A(x)). But you have not followed your own insight. "Only basketballers and footballers are more lazy than every swimmer" means that if something is more lazy then every swimmer then it is a basketballer or a footballer. So your conditional is the wrong way round. Your second try is also wrong because it would mean that for ANY swimmer, someone lazier is a basketballer or footballer. What you want to express is that if someone is lazier than EVERY swimmer then that person is a basketballer or footballer.

Have a go at that. If you are still stuck, write a comment and I'll give a formula for it.

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  • Thank for that breakdown. It really helped. I thought I would have to reference being a footballer or basketballer first but looking back, I have no idea why I rephrased it that way. Aug 5, 2023 at 23:58

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