User talk:Bobby Jacobs

Combinatory logic: I'd keep long versions of I* expressions, and add a footnote that those are the shortest I* which are not I. What do you think? --Blashyrkh (talk) 15:44, 21 June 2026 (UTC)

OK, but technically I is an expression for I*. Bobby Jacobs (talk) 21:41, 21 June 2026 (UTC)

Syntactically they are different combinators, but eta-equivalent. Eta-equivalence is symmetrical relationship: if A is eta-equivalent to B, then B is eta-equivalent to A. There's no non-symmetric relationship between I and I* (and if you believe there is then name it).

Also, it's probably incorrect point of view on combinators, but I mention it anyway (and be known as a heretic): I* postpones effects of x until y is given, while I doesn't. So, they differ in behavior. --Blashyrkh (talk) 04:14, 22 June 2026 (UTC)

Every identity bird is an identity bird once removed, but not every identity bird once removed is an identity bird. Take any identity bird I. Then, for any bird x, Ix=x. Then, for any bird y, Ixy=xy. Since for any birds x and y, Ixy=xy, then I is an identity bird once removed. Therefore, all identity birds are identity birds once removed. However, there might be an identity bird once removed I* that is not an identity bird. It is possible that for any birds x and y, I*xy=xy, but it might not be true that for any bird x, I*x=x. I*x could be a bird that is similar to x in that it responds the same way to any bird y as x, but it is not the same bird as x. Therefore, not all identity birds once removed are identity birds. Bobby Jacobs (talk) 12:33, 25 June 2026 (UTC)

I understand that and agree but not entirely agree. We read the definition I*xy=xy differently. You read it as "I* is a combinator X such that Xxy reduces to xy". Obviously, both I and I* satisfy this definition. And I read it differently: "I* is a rank-two combinator X such that Xxy reduces to xy while neither Xx nor X itself don't reduce any further". In most cases both your and my ways to read a definition produce the same result, except the case when two combinators are eta-equivalent. Not sure who is wrong and who is right, but I think it's rather useless to give I as a definition for I* in the table. --Blashyrkh (talk) 13:09, 25 June 2026 (UTC)

IJ expression for V*

The length-20 expression you just added to Crazy J page - is it the result of bruteforce or have you worked it out by pen and paper? --Blashyrkh (talk) 13:25, 27 June 2026 (UTC)

You already answered here: Talk:Crazy J. --Blashyrkh (talk) 13:55, 27 June 2026 (UTC)

I did it in my head. Let R₁=JR. Then, R₁xyzw=JRxyzw=Rx(Rzy)w=Rzywx=ywzx. It is easy to see that V*=R₁R₁R₁R₁.

R₁R₁R₁R₁xyzw=R₁xR₁R₁yzw=R₁yR₁xzw=R₁zxyw=xwyz

Therefore, V*=J(J(JII))(J(J(JII)))(J(J(JII)))(J(J(JII))). Bobby Jacobs (talk) 13:13, 1 July 2026 (UTC)

Thanks for your contributions to Crazy J

It's really cool to know that your language might be interesting to someone else. --Blashyrkh (talk) 14:33, 12 July 2026 (UTC)

Thanks. Bobby Jacobs (talk) 14:47, 19 July 2026 (UTC)

TMaM, chapter 9, question 11 (A Fact About Kestrels)

"Prove that if a kestrel is egocentric, then it must be hopelessly egocentric." -- R.Smullyan

May you explain it to me? Why the hell the "IF"? Kestrel (K combinator) is not egocentric, because KK is not K: KKxyz = Kyz = y, Kxyz = xz. --Blashyrkh (talk) 14:55, 24 September 2026 (UTC)

if A is false, then "A implies B"=(not A) or B is automagically true. Cleverxia (talk) 00:22, 25 September 2026 (UTC)

Suppose a kestrel K is egocentric. Then, KK=K. Then, for any bird x, KKx=Kx. Since K is a kestrel, KKx=K. Therefore, Kx=K. Then, K is hopelessly egocentric. Bobby Jacobs (talk) 18:26, 27 September 2026 (UTC)

No offense, but it's bullshit. I can prove that K=I from this a priori false statement. Look: suppose K is egocentric: KK=K. Then, KKIIK=KIIK. Since K is a kestrel, KKIIK=KIK=I and KIIK=IK=K. I=K. q.e.d. The same way I can prove that all combinators are equivalent. What's the value of this puzzle? --Blashyrkh (talk) 22:43, 27 September 2026 (UTC)

It is probably used as a basic exercise about kestrels. He later proves in Question 19 that if a kestrel is egocentric, then it is the only bird in the forest. Suppose KK=K. Then, for any birds x and y, KKxy=Kxy. Since KKxy=Ky and Kxy=x, then Ky=x. Then, K is fixated on every bird x. Therefore, K is the only bird in the forest. We could take this one step further with what you said. If KK=K, then for any birds x, y, z, KKxyz=Kxyz. Since KKxyz=Kyz=y and Kxyz=xz, then xz=y. Then, every bird x is fixated on every bird y. Therefore, there is only one bird in the forest. Bobby Jacobs (talk) 14:23, 2 October 2026 (UTC)

Hmmm, it begins to make sense to me. Other way round: suppose that there's only K in the forest. KKK is K by definition of a kestrel, but what is KK? It can't keep silence, it's agains the Rule of The Forest. It must respond, and "K" is the only option (in this forest). So, KK=K. And in this forest the equation KK=K doesn't lead to contradictions: KKK=K both because of K being a kestrel and because KK=K (KKK=(KK)K=KK=K).

Thanks for explanation. --Blashyrkh (talk) 14:54, 2 October 2026 (UTC)