clojure,scheme,lisp,common-lisp , No-argument (and) returns t

## Question:

Tag: clojure,scheme,lisp,common-lisp

Both CL and Scheme define `(and)` to return `t` (or `#t`) with no arguments.

I'm trying to understand the rationale for this. My naive assumption is that an empty set is false, and passing in zero arguments feels like passing in nothing that can be true.

Edit: clojure follows the same convention. I must be missing some basic Lisp assumption.

The empty product is 1. The reason is that 1 is a neutral element for `*`.

If you have the product of 2 and 3 and then multiply by the product of nothing, you will get 2*3*1 = 6. We can write

``````  (product (product 2 3) (product))
= (product 6 1)
= 6
``````

The same computation with `and`:

``````  (and (and #t #t) (and))
= (and #t ?)
= #t
``````

We want the empty `(and)` to give a value `?` that doesn't affect the result. The answer is `#t` since `#t` is a neutral element.

``````(and x #t) = x   for all boolean x
``````

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