What is 2 + 2? Its kinda important. for a very important school project.

I do not know how to solve this incredibly difficult question. The complexity rivals those of the millennial problems. Instead, I’ll let reddit user u/sellopou speak his words.

Take a computational approach to the problem. Specifically, consider the encoding of natural numbers in the simply-typed lambda calculus called Church numerals. In this encoding, the number 2 is represented by a function that looks like this:

2 ≡ λf.λx.f (f x)  

In other words, the number two is a function that takes a function and some value (let's assume that we're comfortable with currying here), and applies the function twice to the value. In general, a number n is represented by a function that takes a function and some value and applies f to the value n times.

Using the Church encoding, you can define operations on the numbers as well. In this case, you're interested in plus, which takes two numbers and returns a number

plus ≡ λm.λn.λf.λx.???  

So what's in the body? Well, the body has to be n + m applications of f to the value x. If we pass f and x to n, we get n applications–call that r, and if we apply m to f and r, that'll be m + n applications of f:

plus ≡ λm.λn.λf.λx.m f (n f x)  

So the expression 2 + 2 using this encoding is:

plus 2 2 ≡ (λm.λn.λf.λx.m f (n f x)) (λf.λx.f (f x)) (λf.λx.f (f x))  

Now β-reduce until you get the result:

plus 2 2 ≡ (λm.λn.λf.λx.m f (n f x)) (λf.λx.f (f x)) (λf.λx.f (f x))  

        ↦ (λn.λf.λx.(λf.λx.f (f x)) f (n f x)) (λf.λx.f (f x))  

        ↦ (λf.λx.(λf.λx.f (f x)) f ((λf.λx.f (f x)) f x))  

        ↦ (λf.λx.(λx.f (f x)) ((λf.λx.f (f x)) f x))  

        ↦ (λf.λx.(λx.f (f x)) ((λx.f (f x)) x))  

        ↦ (λf.λx.(λx.f (f x)) (f (f x)))  

        ↦ (λf.λx.f (f (f (f x  

        ≡ 4.  

But now, let's bust out of this theoretical prison and say we have a number 0 in some representation that we can actually work with, as well as a function add1 that knows how to increment a number in that representation. Well, then we can get a useable representation out of our church numeral, by applying to the numeral to add1 and the value:

4 add1 0 = 4  

(x+3)(x+7)

step-by-step explanation: