### Is any man's parent male?

Nice to know these medieval problems are still a difficulty. Anthony argues that 'every man's parent is male' is false because (if I am understanding him) every man has two parents. OK, then if Aristotelian theory is correct, and if that sentence really does have the logical form 'every A is B', it follows that its contradictory 'some man's parent is not male' is true. But what makes that true? Well, take any man you like, say John. Jean is a parent of his, and she is not male, so 'some man's parent is not male' is true because of John and Jean. But John was any man you like. It therefore follows that 'every man's parent is not male' is true.

But that causes a problem. Assuming that every parent is either male or female, 'every man's parent is not male' implies 'every man's parent is female'. But the whole argument began with the assumption that 'every man's parent is male' is false because every man has two parents. Why, by equal reasoning, isn't 'every man's parent is female' false also?

But that causes a problem. Assuming that every parent is either male or female, 'every man's parent is not male' implies 'every man's parent is female'. But the whole argument began with the assumption that 'every man's parent is male' is false because every man has two parents. Why, by equal reasoning, isn't 'every man's parent is female' false also?

Labels: logic, quantification

## 10 Comments:

>> But John was any man you like. <<

No, he isn't! John is John, a constant not a variable. To argue from Fa to ∀x.Fx is quite invalid. You need to start from an arbitrary x and show Fx is true. In this case, to show that an arbitrary choice of (man's parent) is not male. It's not clear from the context whether we are considering some limited set of men and parents (John, Jack, Joe, Jane, Jean, Jackie, ...), in which case it might be true that every choice of (man's parent) is female, but for men and parents in general, and I think this is where Anthony starts, some choice of (man's parent) will be male.

But there is a second fallacy here. Jean shows that 'some (man's parent) is not male' is true. The quantification is over the class (man's parent). Then you move to

>> But John was any man you like. It therefore follows that 'every man's parent is not male' is true. <<

This is a shift to the parsing ((every man)'s parent), ie, to quantification over men. The donkey question is a tissue of ambiguities that needs to be resolved before we can decide whether Aristotelian logic is applicable to it. What did the medievals have to say?

FWIW, I did not argue that "every man's parent is male" is false. I argued that it is not a true sentence.

I don't think every collection of words can be categorized as "true" or "false".

>>No, he isn't! John is John, a constant not a variable. To argue from Fa to ∀x.Fx is quite invalid.

There is a well known principle called 'arbitrary object'. Take any object you like, a. As long as it really is 'any object', i.e. without regard to any other features, then if you can prove Fa, then infer (x) Fx.

Arbitrary objects.

Ed,

I continue to be impressed with what mathematical logicians can come up with! The paper talks about doing mathematics over Fraenkel-Mostowski set theory (FM) rather than the conventional Zermelo-Fraenkel. Top of the second page:

"The FM universe contains a range of extra elements that can be regarded as the denotations of variables. That is, using FM set theory we can interpret variables directly in the universe (as arbitrary objects), without making any recourse to variable assignments as in the familiar Tarski truth definition."

What, in our current example of men and parents, corresponds to these extra elements?

Your argument would show that all primes are odd. After all, 17 is odd and I chose it arbitrarily, didn't I? A valid use of ∀-introduction would start with an unknown prime, x. We would have to show that x must be odd assuming only that it is prime, not from knowing which prime it is, And of course we can't do this because 2 is prime and it's even.

In any case, if "every man's parent is male" is of the form "every A is B" (*), then the A is "man's parent", not "man". So the argument would have to go "Take any man's-parent you like..." And an arbitrary man's-parent might be male, and might be female.

(This is what David explained, differently, and probably more eloquently, when he discussed quantification. But as you seem to have ignored that part I'm repeating it using my own words.)

(*) And I would argue that this is *not* a natural way to read the sentence, though I suppose it is a conceivable reading ("every parent-of-a-man is male"). Under this interpretation, the sentence would be false (and surely you agree that "every parent-of-a-man is male" is false).

>>Your argument would show that all primes are odd. After all, 17 is odd and I chose it arbitrarily, didn't I?

No, you chose an odd number rather than any number. The property that you prove must be independent of the selection process.

Actually, I chose a

prime'at random'. But that isn't good enough, because the subsequent proof must be valid for all possible choices, and it won't work for the prime 2. Returning to the original example1. Jane is John's parent and she is female

2. Jean is Jack's parent and she is female

3. Jackie is Joe's parent and she is female

... and so for every man

Thus, every man's parent is female.

much depends, as I suggested earlier, on what is hidden in the ellipsis. If it is empty (ie, there are just three men each having a female parent) then it is indeed the case that, in this context, every man's parent is female. Proof: by enumeration. But suppose the ellipsis hides the premiss

23. Jim is Jerry's parent and he is male.

We can no longer argue that a parent chosen arbitrarily is female. For we might choose Jerry's parent Jim. As you rightly say, the proof must be independent of the selection process. The easiest way of ensuring this is to construct a proof without knowing what the selected element is.

OK but assume Jemima is Jerry's mother. Then Jemima and Jerry together verify "some man's (i.e. Jerry's) parent (Jemima) is female.

And if we can verify 'Some A is B' for any A whatsoever, why shouldn't that demonstrate 'every A is B'?

>>In any case, if "every man's parent is male" is of the form "every A is B" (*), then the A is "man's parent", not "man".

Actually you are right.

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