Etl (pronounced like "nettle" without the n) is an esoteric programming language for transforming expressions. There are three data types in Etl: pairs, varaibles and operators. Pairs are pairs of the said datatypes. Pairs are written with parenthesies and arguments seperated by spaces. Operators start with a ! which is followed by any sequence of non-whitespace characters. Variables are all other sequences of characters. Here is an example:

(hello (((!0) foo) bar))

Programs are written as a set of replacement rules and a list of expressions to apply the rules to. Here is the syntax and how they work:

rule swap (!S (a b)) -> (b a)
(!S ((!S (a b)) c))

which outputs (c (b a)).

The general syntax for rules is:

rule <name> <pattern> -> <replacement>

where <name> is replaced by a string of the name of the rule (currently there is no use for the name, one will be added in the future), and <pattern> and <replacement> are replaced by expressions.

First we do a process called pattern matching which may or may not yield a mapping between the dummy names and actual expressions called "bindings". For example when we match the expression (!S (a (b c))) against the pattern (!F (a b)) we get no match because the operator !F doesn't have the same name as the operator in its place in the expression. However if we have the exprssion (!S (x (y z))) it matches producing the following bindings:

{
"a": x
"b": (y z)
}

Now that we have those bindings we can produce a different expression. That's what the replacement part of the rule is for. Say that we have a rule

rule swap (!S (a b)) -> (b a)

and an expression

(!S (x (y z)))

Our rule is applied recursively outside-in. So first the program generats the outermost bindings:

{
"a": x,
"b": (y z)
}

Then the language takes the replacement expression and replaces every instance of "a" and "b" with the corresponding binding. In the end of evaluation our expression we end up with

((y z) x)

What happens when you have two conflicting rules? What happens when there are variables in the replacement expression not found in the pattern? I don't know! Experiment and have fun. This shouldn't be defined in the specification of the language because I am too lazy to research it.

Currently the programming language is only available as a repl. To exit it simply type bye.

rule r1 (!R ((x y) (z))) -> (!R ((x) (z y)))
rule r2 (!R (!0 x)) -> (x)
(!R (((((!0 a) b) c) d) !0))

outputs:

((((!0 d) c) b) a)

https://en.wikipedia.org/wiki/SKI_combinator_calculus

rule S (((!S x) y) z) -> ((x z) (y z))
rule K ((!K x) y) -> (y)
rule I (!I y) -> (y)