But since you didn't specify lambda calculus in your question, but instead tagged both Racket and Scheme, I will give you an answer that works there.

Consider if you have true and false defined as above. Then writing not is as simple as:

(define not
(lambda (x)
(cond [(equal? x true) false]
[(equal? x false) true]
[else (error "Value is not true or false")])))

This has the same semantics listed above: ((not true) 1 2) gives 2 and ((not false) 1 2) gives 1. But it has the added benefit that you can actually directly examine it to see the value, rather than just applying it to another procedure. For example:

This leads to a whole discussion of PL theory about what can and cannot be decided with the lambda calculus. That I will elide here. But if you are interested, I recommend you check out this book, or that book.

So, @PetSerAl is absolutely right that the correct answer to this (if we are looking for a lambda-calculus-esque solution) is:

Thus:

`((not true) 1 2)`

gives`2`

and`((not false) 1 2)`

gives`1`

.But since you didn't specify lambda calculus in your question, but instead tagged both Racket and Scheme, I will give you an answer that works there.

Consider if you have

`true`

and`false`

defined as above. Then writing`not`

is as simple as:This has the same semantics listed above:

`((not true) 1 2)`

gives`2`

and`((not false) 1 2)`

gives`1`

. But it has the added benefit that you can actually directly examine it to see the value, rather than just applying it to another procedure. For example:You can even now use

`equal?`

(or really`eq?`

on them), whereas before you would have always gotten`#f`

if you tried that:This leads to a whole discussion of PL theory about what can and cannot be decided with the lambda calculus. That I will elide here. But if you are interested, I recommend you check out this book, or that book.