The question was: which symbolization best fits the sentence
No matter who you pick, there's someone who is a wife if and only if that person is a husband.
These were the possibilities:
a) ∀x∃y(Wy ↔ Hx) b) ∀x(∃yWy ↔ Hx)
So far, one person thinks b) is better and another doesn't see a difference.
What if I told you that one of them, if true, implies that either everyone is a husband or no one is? Because that's actually true. In that case, would you say that the one with this consequence IS the best symbolization? or ISN'T? And which one has this implication in any case?
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7 comments:
I think it's "A" because I think "B" fits your description of either everyone is a husband or no one is and I don't think that would be the best fit?
I think they're both the same:
@x$y(Wy = Hx) <=> $y(Wy = Ha) => We = Ha
@x($yWy = Hx) <=> $yWy = Ha => We = Ha
they seem to be equivalent:
@x$y(Wy = Hx) = @x($yWy = Hx)
I will attempt a proof of the sentence above to see if it fits
I think they are the same because the $y distributes to the Wy and the @x distributes to the Hx. In either sentense, they distribute appropriately. So, I think it means the same thing in a) and b).
I think a and b are the same just written out in a different ways. C Seitz
I think that B has the implication that everyone is a husband or no one is because of the scopes of the parentheses. I think this is the best symbolization for the one with this consequence because I interpret the sentence to read "[There's someone who is a wife] iff that person is a husband," and that is how the parentheses in B break up the sentence.
I believe A is better than B. I think that having the existential quantifier on y inside the parens is tricky since it is in an IFF and in this case I don't think the intention is to tie the existence of at least one wife with the assertion that all x are or are not husbands. In other words- Doesn't B effectively say something like "either unless every x is a Husband then there are no wives at all, or no x is a husband" ?
That's good fun, pity I came across it only just now.
Obviously both a) and b) are valid interpretations of the written sentence, showing that that sentence is simply ambiguously formulated (something that is common in natural language). But a) and b) are not equivalent.
b) says two things must be true about any chosen x:
if x is a husband, then there must be a wife (this is simply a consequence of the definition of "husband" and thus "always" true); and if x is not a husband, there are no wives at all (and hence no one is a husband). The latter is clearly violated in our actual world of people: there are (lots of) non-husbands and (yet) there are (lots of) wives.
a) says: for any chosen x there is a y whose property of being a wife depends on x having or not having the property of being a husband. Now, as these properties are fixed at any time (either present or absent), we can prove a) to be true in our actual world of people: pick x; if x is a husband, pick y to be his wife, then Wy↔Hx is "trivially" true; on the other hand, if x is not a husband, pick y to be someone who is not a wife (which is feasible in our world) - then again Wy↔Hx is "trivially" true.
To carry this a little further, let's consider formula a) as such, with H and W being any properties that objects in a given world may or may not have. Above we noted that the proposition A↔B was "trivially" true in the given situation simply since either both A and B were true or both A and B were wrong. Now, in a world of quantum objects there is a less trivial way of having A↔B true. We may in fact limit the "world" to one that consists of pairs of entangled particles where all the pairs are independent of all other pairs. Then for any individual particle there is a unique entangled partner and the entanglement means that there are properties, call them H and W, such that if partner 1 is found to have H then partner 2 is found to have W; the difference to the situation with husbands and wives is that our partner particle 1 can be found equally well not to have H - in which case partner 2 is necessarily also found not to have W. The properties H,W are contingent and not determinate until they are measured, but between each pair of partners the properties are perfectly correlated.
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