Church zero always returns the identity function, regardless of its first argument. In other words, the first argument is not applied to the second argument at all.

Church one applies its first argument f just once to its second argument x, yielding f(x)

Church two applies its first argument f twice to its second argument x, yielding f(f(x))

and each successive Church numeral applies its first argument one additional time to its second argument, f(f(f(x))), f(f(f(f(x)))) ... The Church numeral 4, for example, returns a quadruple composition of the function supplied as its first argument.

Arithmetic operations on natural numbers can be similarly represented as functions on Church numerals.

I have written the following functions:

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```
(define (zero f x) x)
(define (uno f x) (f x))
(define (due f x) (f (f x)))
(define (tre f x) (f (f (f x))))
(define (quattro f x) (f (f (f (f x)))))
(define (cinque f x) (f (f (f (f (f x))))))
(define (sei f x) (f (f (f (f (f (f x)))))))
(define (sette f x) (f (f (f (f (f (f (f x))))))))
(define (otto f x) (f (f (f (f (f (f (f (f x)))))))))
(define (nove f x) (f (f (f (f (f (f (f (f (f x))))))))))
(zero inc 0)
;-> 0
(uno inc 0)
;-> 1
(due inc 0)
;-> 2
```

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```
(setq f inc)
(setq x 0)
(zero f x)
;-> 0
(sei f x)
;-> 6
```

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```
(define (succ n f x) (f (f n x)))
(succ 0 inc 0)
;-> 1
(succ 3 inc 0)
;-> 4
(succ 2 inc 0)
;-> 3
```

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```
(define (plus m n f x) (f m (f n x)))
(plus 3 2 inc 0)
;-> 5
(plus (due inc 0) 5 inc 0)
;-> 7
(plus (due f x) 5 f x)
;-> 7
```

1) multiplication (mult)

2) predecessor (pred)

3) subtraction (minus)

Can anyone help me?

cameyo

p.s. I'm not sure if what I wrote is correct

p.p.s. I forgot to translate the numbers into English :-)