The Luby-Rackoff theorem states that if a round function is a secure pseudorandom function (PRF) then 3 rounds are sufficient to make the block cipher a pseudorandom permutation (PRP).
PRPs are invertible whereas PRFs are not. How come 3 rounds of a PRF will make an invertible block cipher out of a non-invertible function?
The proof is loosely as below.
Lets assume a one round Feistel network, where $2n$ bits are divided into $n$ bits each $L_0, R_0$
The encryption is defined as
$L_{1} = R_{0}, \\ R_{1} = L_0 \oplus f(R_0) $
where f is any random function (PRF) and $\oplus$ is XOR operation
Now the cipher text is $L_{2} = R_{1}, R_{2} = L_1 $
Decryption is same as encryption circuit as defined above.
The input to decryption is $L_2, R_2$.
So decryption is defined as below
$ L_{3} = R_{2} \\ R_{3} = L_{2} \oplus f(R_2) $
Where the plain text should be considered $R_3, L_3$
Now lets substitute from $ L_{3} = L_{1}, \\ R_{3} = R_{1} \oplus f(L_1) $
Now lets substitute further to get the plain text is $R_3, L_3$ which is $L_0, R_0$ as shown below $ L_{3} = R_{0}, \\ R_{3} = L_0 \oplus f(R_0) \oplus f(R_{0}) ,\\ R_{3} = L_0 $
So it does not really matter if $f(R_0)$ is reversible or not. And the same holds good for any number of rounds.
There is also an intuitive explanation here
