Talk:Combinatory logic

Simple Bases for Combinatory Logic

It might be worth mentioning that CL (combinatory logic) can be defined in the KS-basis as follows ...

Syntax
------
<combinator> ::= K | S | ( <combinator> <combinator> )

Semantics
---------
((K x) y)  -->  x
(((S x) y) z)  -->  ((x z)(y z))

That's all! And even the right-parenthesis is not needed for disambiguation -- e.g. leave it out altogether, and replace '(' with '*':

Syntax
------
<combinator> ::= K | S | * <combinator> <combinator> 

Semantics
---------
* * K x y  -->  x
* * * S x y z  -->  ** x z * y z

One further step, replacing the KS-basis with any single-combinator basis (call it B -- there are many single-combinator bases), brings us to an even more minimalistic language (we get the language Iota by taking B = i = \x.xSK):

Syntax
------
<combinator> ::= B | * <combinator> <combinator>

Semantics
---------
* B x  -->>  <def'n of the basis combinator B>   (e.g. * i x -->> * * x S K)

It can't get much simpler!

Added note:

It's a theorem in CL that every single-combinator basis is improper; that is, a single combinator cannot properly serve as a basis, because it is itself necessarily defined in terms of two or more other combinators. E.g., the i in Iota is defined in terms of K and S (which are proper combinators -- contrast \x.xSK with \xy.x etc.). So although programs in the resulting language are binary strings on the alphabet {*,B}, they actually amount to encodings of trinary strings on the alphabet {*,K,S}.

--r.e.s.

Role of free and bound variables

In the content page on Combinatory Logic, of course you mean to say

"Each combinator is like a function or lambda abstraction, but without any FREE variables."

--r.e.s.

Yes, thank you. --Chris Pressey 00:39, 2 Jul 2005 (GMT)

Don't understand something

I don't understand something about this article. For example "S K K x". I have understood that "x" also is a combinator, so it's also a series of S and K symbols. Say x is a K. Then what is the meaning of this last K? There's nothing behind it anymore, so it isn't applied to anything. There's no possible thing that x can be that doesn't require another combinator behind it. But there has to be a last combinator somewhere! --Aardwolf 02:21, 27 Oct 2005 (GMT)

You don't need to supply combinators with 'arguments' - when no more reductions are possible, they remaining combinators just sit there. So KKS reduces to K, and that's it. --Safalra 10:29, 27 Oct 2005 (GMT)

Keywords

The keyword pointfree programming (or point-free) should probably appear somewhere on the wiki, because they apply to these combinatory logic systems like the ones on this page and Unlambda, and also for some stack-based languages with no variable bindings such as Underload. I don't know where to mention this, perhaps there should be a separate article for the pointfree programming concept. --(this comment by B jonas at 08:03, 13 September 2018‎ UTC; please sign your comments with ~~~~)

Is there anything between IJ and complete bases?

IJ basis contains all non-cancellative combinators (so called Aristocratic system). Is it enough to add any cancellative combinator to IJ basis to make the basis complete? In other words, is there a non-complete system (other than Aristocratic system itself) that includes Aristocratic system as a subset? --Blashyrkh (talk) 23:14, 23 January 2026 (UTC)

This is a great question. I think that the answer is yes, but I don't have a proof. Certainly IJK is complete. Corbin (talk) 00:11, 24 January 2026 (UTC)

An example proves nothing, but...

I've checked first 23 37 cancellative combinators (User:Blashyrkh/Between IJ and SK), and in all cases an expression of K combinator in IJX basis exists (i.e. IJX basis is proven to be complete where X is the cancellative combinator in question). --Blashyrkh (talk) 08:11, 24 January 2026 (UTC)

What's the exact definition of cancellative combinator?

The page says:

A combinator is cancellative if it does not use all of its arguments.

Should we consider λx.xK a cancellative combinator? Pro: it can't be expressed in IJ basis. Con: it uses all its arguments (x). --Blashyrkh (talk) 12:00, 24 January 2026 (UTC)