How to reduce Untyped λ-Calculus to Normal Form?

I have an assignment to do which involves reducing an Untyped λ-Calculus expression to Normal Form. I am struggling to come to terms with Lambda Calculus though.

For example, one small part of the expression is:

((λ n f x . n f (f x)) (λ f x . x))

Does this mean that we substitute (λ f x . x) in for every n, f and x? Giving us the following:

((λ n f x . (λ f x . x) (λ f x . x) ((λ f x . x) (λ f x . x))))

If so, can this be reduced further? I know I am probably way off with this, but I just do not understand it; any help would be appreciated.

EDIT: Thanks to @ljedrz Comment.


Before doing the reduction is worth noting that:
$$ (\lambda n f x . n ~ f (f ~ x)) = (\lambda n f x .((n f) (f ~ x))) $$
This is a convention, the parenthesis always associate to the left

It can be reduced further, but I advise you to change the second expression to an $\alpha$-equivalent one.

$$ \Big(\big(\lambda n f x . (n ~ f) (f ~ x) \big) ~ (\lambda h y . y) \Big) $$

This might not be necessary in this specific problem, but might also save you some headache later trying to keep track if some free variables got captured.

\begin{align*}
\Big(\big(\lambda \color{red}{n} f x . \color{red}{n} ~ f (f ~ x)\big) ~ \color{red}{(\lambda h y . y)} \Big) &\rightarrow_\beta \big[n ~ / ~(\lambda h y . y)\big] \big(\lambda f x . n ~ f (f ~ x)\big)\\
&\rightarrow_{\alpha} \big( \lambda f x. (\lambda h y. y) ~ f ~ (f~x) \big)\\
\Big( \lambda f x. \big((\lambda \color{red}{h} y. y) ~ \color{red}{f}\big) ~ (f~x) \Big) &\rightarrow_{\beta} \Big(\lambda f x.\big[h ~ / ~ f \big]\big(\lambda y.y\big) ~ (f ~ x) \Big)\\
\Big( \lambda f x. ((\lambda y. y)) ~ (f~x) \Big) &\rightarrow_{\beta} \Big(\lambda f x.\big[y ~ / ~ (f ~ x)\big]y\Big)\\
&\rightarrow_{\alpha} \Big(\lambda f x. (f ~ x) \Big)
\end{align*}

Which cannot be reduced further.

StackExchange Computer Science Q#86615, answer score: 3