Does the position of ∃ matter?

Question

Everyone who makes an A aces some final exam

Is this

∀x,∃y: makes(x, A) => aces(x, y)

the same as

∀x: makes(x, A) => ∃y: aces(x, y)

where x represents everyone and y represents exams.

Does the position of ∃ matter?

In this case both are equivalent, however, it is not the case for ∀x,∃y P(x,y) and ∃y,∀x P(x,y) — hhsaffar, Nov 14 '13 at 21:59

score 1 · Accepted Answer · answered Nov 14 '13 at 21:58

1

No, they are equivalent.

This is part of a theorem called "Prenex normal form".

answered Nov 14 '13 at 21:58

Shay Ben Moshe

1,070

Did · Answer 2 · 2013-11-14T21:56:53.293

0

Let $M=\{x\mid\text{makes}(x,A)\}$ and $E$ the set of exams. Then, both assertions are equivalent to: $$ \forall x\in M,\quad\exists y\in E,\quad\text{aces}(x,y). $$

edited Nov 14 '13 at 21:56

answered Nov 14 '13 at 21:51

Did

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Does the position of ∃ matter?

2 Answers2