# Exercise 1.6

Alyssa P. Hacker doesn't see why if needs to be provided as a special form. ''Why can't I just define it as an ordinary procedure in terms of cond?'' she asks. Alyssa's friend Eva Lu Ator claims this can indeed be done, and she defines a new version of if:

```
(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
(else else-clause)))
```

Eva demonstrates the program for Alyssa:

```
(new-if (= 2 3) 0 5)
5
(new-if (= 1 1) 0 5)
0
```

Delighted, Alyssa uses new-if to rewrite the square-root program:

(define (sqrt-iter guess x)
(new-if (good-enough? guess x)
guess
(sqrt-iter (improve guess x)
x)))

What happens when Alyssa attempts to use this to compute square roots? Explain.