lambda calculus with church numerals

Question

today I found this term in our exercises: ((^fx.f(f(f x)) ^gy.g(g y )) ^z.z + 1) (0)

I am quit unaware how to solve this type of question. I know this is the church numeral 3 , 2 , the identity function +1 and zero. But for my understanding it lacks an operator like +-*/ as ex.

(λmn.m (succ n)) to connect two church numerals?

Some hints will be greatly appreciated.

Answer 1

Your term is the application $3\ 2\ succ\ 0$, where $succ$ is the successor function.

If you task is to reduce this term to a beta normal form:

First we can observe that for terms $M$ and $N$, the application $MN$ beta-reduces to the $M$-fold application of $N$, which is

• in the case of $N$ an operation: the $M$-fold application of the operation $N$ to whatever term $P$ the term $MN$ is applied to; i.e., $MNP$ is $\underbrace{N(N(...N(}_{M\text{ times}} P)))$,
• in the case of $N$ a church number: the (church number representation of the) exponentiation of $N$ with $M$; i.e, $MN$ with $M,N$ numbers is $N^M$.

as you can easily verify with simple numbers like $2$ and $3$.

So $3\ 2$ is $2^3$, which beta-reduces to $8$, and $3\ 2\ succ$ creates an 8-fold application of $succ$, which is then applied to $0$.
So your term is the $2^3$-fold = $8$-fold application of the $succ$-function to the number $0$, that is, $\underbrace{succ(succ(...succ(}_{8 \text{ times}} 0)))$, which eventually beta-reduces to the church number $8$, $\lambda fx.f(f(f(f(f(f(f(f(x))))))))$.

Answer 2

Essentially your term is the application $3\ 2\ {\sf succ}\ 0$. So we start performing beta reduction steps. Since $3\ f\ x = f(f(f x))$, we get $2\ (2\ (2\ {\sf succ}))\ 0$. From here, we get $(2\ (2\ {\sf succ}))\ ((2\ (2\ {\sf succ}))\ 0)$. And so on -- I'll leave the steps to you.

If I see correctly, after a long-ish sequence of beta steps we should reach the numeral $8$ as a normal form. Try checking that $(2\ (2\ {\sf succ}))\ 0 = 4$, and after that check that $(2\ (2\ {\sf succ}))\ 4 = 8$.