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$$∀x∀y ((\text{Cube}(x) ∧ \text{Cube}(y)) → x = y)$$

To me, this statement means that for all cubes x and y, if you pick up a cube from the box, it will always be the same cube.

I understand this to mean that there is exactly one cube in the box.

I don't understand how the statement means that there is at most one cube in a box.

ryang
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iSuckatMath
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    Think about what happens if you replace "Cube" with "Unicorn." At no point have you asserted that instances of the predicate in question actually exist, merely that we can't have two different ones. (I've left this as a comment as opposed to an answer since I suspect this is a duplicate question.) – Noah Schweber Nov 29 '21 at 01:27
  • To say that there is exactly one $x$ such that $P(x)$, you need to say not only that there is at most one such $x$, as your formula states, but also that there's at least one: $... \land, (\exists x)P(x)$. – BrianO Nov 29 '21 at 01:57
  • $\exists x$ means "at least one". Thus, "exactly one" will be: "at least one and no other", that means that we have an object $x$ satisfying the specified condition and that every object $y$ satisfying the condition will be identical with $x$. – Mauro ALLEGRANZA Nov 29 '21 at 09:27
  • So, what about "at most"? We have to remove the "at least one" part, because "at most" is compatible with zero. – Mauro ALLEGRANZA Nov 29 '21 at 09:28
  • In conclusion: "at least one Cube" = $\exists x \text {Cube}(x)$. "exactly one Cube" = $\exists x [\text {Cube}(x) ∧ \forall y (\text {Cube}(y) → x=y)]$. "at most one Cube" = $∀x∀y[(\text {Cube}(x) ∧ \text {Cube}(y))→x=y]$ – Mauro ALLEGRANZA Nov 29 '21 at 09:29

1 Answers1

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  • $$∀x∀y\;\big((\text{Cube}(x) ∧ \text{Cube}(y)) → x = y\big)$$

    Yes the above statement allows for the possibility of exactly one cube in the box, as you have explained.

    But it also allows for the possibility of no cube in the box: in this case, whichever pair of objects you pick, $(\text{Cube}(x) ∧ \text{Cube}(y))$ is false and the entire statement consequently vacuously true.

    Here's an equivalent, more succincct formulation of at most one: $$\exists x\forall p\;(\text{Cube}(p)\to p=x).$$

  • In contrast, exactly one cube can be translated as $$\exists x\forall p\;(\text{Cube}(p)\leftrightarrow p=x).$$

ryang
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  • i dont understand how it being vacuoulsy true means that there is zero cubes. wouldn't that just me the cube and object x are equal to each other, which would be a contradiction. – iSuckatMath Dec 02 '21 at 01:23
  • @iSuckatMath 1. i dont understand how it being vacuoulsy true means that there is zero cubes I said "consequently vacuously true"; by reversing my implication direction, you have misread and misunderstood my explanation. 2. wouldn't that just me the cube and object x are equal to each other To avert needless confusion, please think of $x$ not as being equal to a cube, but as having the property of being a cube [Sweet(nectarine) means "nectarine is sweet", but Sweet=nectarine doesn't make sense]. – ryang Dec 02 '21 at 04:47
  • "whichever pair of objects you pick" means "for all x and y"; "Cube($x$) ∧ Cube($y$) is false" means "the statement's left side is false"; "and the entire statement consequently vacuously true" means " this makes the whole statement true" and does not mean "this makes the statement's right side $x=y$ true". To be super clear: the statement in its entirety is an implication and, by definition, is true whenever its left side is false.
    • – ryang Dec 02 '21 at 04:55