Thursday, August 2, 2012

Deontologists are lazy consequentialists and consequentialists are jaded deontologists

Discuss.

He thinks he's so clever... but he's really just a deontologist at heart.

I think it would have looked much beter if they left the head on.

1. Dr. Michael Emmett Brady, who was a triple major in Philosophy, Mathematics, and Economics, opposes Jeremy Bentham's ethical system and its mathematical derivation (by Frank P. Ramsey and Leonard J. Savage) in the form of Subjective Expected Utility.

Adam Smith and John Maynard Keynes both disagreed with Jeremy Bentham's hedonic calculus, and may have some overlap in their decision theories. Gavin Kennedy has reviewed Dr. Michael Emmett Brady's paper before, and seems to support Dr. Michael Emmett Brady's conclusions.

http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1728225

2. Bentham saw it as his duty to encourage people to look towards the happiness of themselves and others. I'm not sure there is anything contradicting consequentialism there. I wonder if he enjoyed it, I expect he did.

3. Let me see if I follow what you're saying.

Let X = Deontologist and Y = Consequentialists. f() shall be the lazy function; g() is the jaded function.

X = f(Y)
Y = g(X)

Thus, (g ∘ f)(X) = g(f(X)) = X

We can state that (g ∘ f) and (f ∘ g) are identity function compositions that furthermore possess commutativity (at least under the given conditions), and thus (g ∘ f) = (f ∘ g). So Lazy, Jaded people are the same as Jaded, Lazy people.

The immediate consequence of this is that Deontologists are Lazy, Jaded Deontologists, but then this can be further unfolded to Deontologists are Lazy, Jaded, Lazy, Jaded, Lazy, Jaded Deontologists. All Consequentialists are also Lazy and Jaded in equal proportion.

By using the equivocative property, we can state:

∴ Consequentialists = Deontologists

...and they're all just awful.

Q.E.D.

4. It has been brought to my attention by a friend (who is more math-mature than I) that if we are willing to limit our domain a bit more, we can state this equality more robustly.

f(Y) = X+1
g(X+1) = Y

If g(X+1) = g(X)+1 and f(Y-1) = f(Y)-1, then:
g(X) = Y-1
f(g(X)) = f(Y-1) = f(Y)-1 = X
therefore g(X) = X

So, under the above conditions, f() and g() can both be shown to be identity functions, and therefore X = f(Y) = Y.

Not sure what that floating +1 would be. I guess Lazy (slightly smarter Consequentialists) = Lazy (Consequentialists) who are slightly smarter? But that may not conform to the way I've been carrying this out.

1. This is the most enjoyable comment I've read in a long time..

5. (Or maybe he was making a tautology joke)

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