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What happens with the output of the round function and the previous half block? Nearly every Feistel networks XOR them. What happens if you XOR any block with an all "0" block? You get the previous block. Example in binary notation with 8-bit blocks: $$\text{0100 1100} \oplus \text{0000 0000} = \text{0100 1100}$$ So even with a million rounds the output ...
If each round function outputs all $0$'s then if $r$ is an even number you get back original string. If $r$ is an odd number you just swap the $(L_0, R_0)$ components of the input. XOR basically has no effect on the string. Just follow the proof given in this answer Luby-Rackoff theorem confusion