Horse of another color.

[PAULK]

Inspired by "All cats have ten tails." AND The Wizard of Oz, in which the guard at Oz admits Dorothy after she announces she is here to see the wizard, saying "Why didn't you say so? That's a horse of another color."

And there is no attempt to misinterpret 'no cat' or, in this case, 'no horse.'


There is no horse of a different color. Proof by induction;

Theorem: Every set of n horses contains horses of only one color.(i.e. never more than one color, with the understanding that a 'paint' horse is a single color.)

Base case: n = 1. Clearly a set containing only one horse cannot have horses of more than one color.

Assumption: n = k. Every set of k horses contains horses of only one color.

To prove: n = k + 1. Every set of k+1 horses contains horses of only one color.

Given a set of k+1 horses. Put all the k+1 horses inside a corral. Remove one of the horses -- call him Araby. What remains in the corral is a set of k horses which, by assumption, are all the same color, say, Chestnut. Now restore Araby to the corral and remove a different horse -- call him Benjy. You again have k horses in the corral, which must be the same color, so Araby is the same color as the others, i.e. Araby is also a Chestnut. So all the horses are Chestnut and the k+1 horses are all the same color.

 

[Denis]

I believe you.

 

[daon]

Betcha can't prove the n=2 case.

 

[tkhunny]

From a logical point of view...I don't like the trivial case n = 1. There is not actually a comparison with only one item in the set, so the word "different" has no meaning. You must start with n = 2, which you cannot.

From a Wizard point of view... As I recall, that particular horse was a different color every time it appeared on the screen, suggesting even a single horse may be a different color, depending on when you look at it. You have assumed color stability. Horses never sunburn?

 

[PAULK]

Betcha can't prove the n=2 case.

I guess I can't fool you guys.

 

[PAULK]

From a logical point of view...I don't like the trivial case n = 1. There is not actually a comparison with only one item in the set, so the word "different" has no meaning. You must start with n = 2, which you cannot.

>> No, I think it is correct to start with n=1. 'No horse of another color." is properly interpreted to mean "There do not exist at least two horses in the set with different colors." But as to your complaint that I cannot prove it for n = 2, THAT's a horse of another color, which, as you see, does not exist.

From a Wizard point of view... As I recall, that particular horse was a different color every time it appeared on the screen, suggesting even a single horse may be a different color, depending on when you look at it. You have assumed color stability. Horses never sunburn?

>> You're right! I remember that, now. Lots of cute stuff in that movie.

 

[daon]

From a logical point of view...I don't like the trivial case n = 1. There is not actually a comparison with only one item in the set, so the word "different" has no meaning. You must start with n = 2, which you cannot.

>> No, I think it is correct to start with n=1. 'No horse of another color." is properly interpreted to mean "There do not exist at least two horses in the set with different colors." But as to your complaint that I cannot prove it for n = 2, THAT's a horse of another color, which, as you see, does not exist.

Well. With this logic one could prove that an equivilance relation need only symmetry and transitivity. Reflexivity would be free...

Assume a~b. Then b~a. But a~b and b~a => a~a. There is not necessarily a 'b', different from your 'a', equivilant to your 'a'.