*Keith Douglas*

Nothing to analyze from last time. Reader Alex mentions some analytical categories to help out and also mentions existing legal categories in Canada — thanks!

**Curry’s Paradox**

This is the paradox whereby curry always seems hotter when you make it yourself! No, actually, the paradox is due to the expounder of what is called combinatory logic, and a Haskell Curry. (This is not to be confused with Paul Curry’s missing-square paradox, which is harder for me to reproduce here but can be viewed on YouTube.) This is the Curry for which currying (and Haskell) are named in computing.

Consider the following proof (adapted from Jeremy Avigad’s *Mathematical Logic and Computation*).

1. Let *A* be the sentence, “If *A* is true, then Zoroastrianism is the correct religion.”

2. Suppose (for conditional proof) that *A* is true.

3. Then what *A* says is true — i.e., if *A* is true, then Zoroastrianism is the correct religion.

4. Since we are in the scope of the conditional proof, we can conclude that Zoroastrianism is the correct religion.

5. Therefore we have shown, “If *A* is true, then Zoroastrianism is the correct religion.”

6. But that’s just *A*. So we have shown that *A* is true.

7. So then, Zoroastrianism is the correct religion. (“*A* is true” is the antecedent of a conditional, so its consequent follows by modus ponens.)

Ready to convert? Not so fast. Replace “Zoroastrianism is the correct religion” with “The PopolVuh is the world’s holiest text.” Uh oh. Try again with “Zoroastrianism is NOT the correct religion.” Even bigger problem. What is going on here?

I’m not a philosopher, and I might seem a bit naive, but it seems that, by the time we get to statement #3 above, we have a condition within a condition. Is that the problem?

I sketched this out in formal logic but got stuck at step 5. I think it goes something like this:

1. A=(A->Z) (premise)

2. A (premise)

3. A->Z (from 1,2)

4. Z (from 2,3)

5. A->Z (why?)

6. A (from 5,1)

7. Z (from 5,6 by modus ponens)

Step 5 seems to simply repeat what we already proved at step 3, except the grounds seem to be subtly different. 3 follows from 1 and 2. 5 seems to follow from 2 and 4, i.e. we have proved that A entails Z by demonstration (except that 4 depends on 3, therefore circular?). Something like that?

I think the point is that, by allowing A to refer to itself, we create a circularity that allows us to prove any proposition whatsoever. Correct?

Let’s just test this with numbers

A=2 or A = .f. ( or any value that you want to test like sq rt of -1 )

A ( 2 then Zoro is the true religion )

A ( .f. then Zoro is the true religion)

A can never be true since the value of A can’t equal a modified value of itself

I’m gong to take another stab at this, using the disjunctive equivalent of the implication:

A-> Z = not-A v Z (where ‘v’ is the OR operator)

So the premise can be rewritten as:

A = (not-A v Z)

If A is true, then the first disjunct is false, thus the second one (Z) must be true by disjunctive syllogism. QED.

This seems a close cousin to proving any proposition from a contradiction.