Within Wolfram's 1D 2-state cellular automata rules, there are three bits (which we'll call bit_n-1, bit_n, bit_n+1) at t-1 that determine the state of bit_n at t-0. This means there are 2^3 = 8 ways to determine the state of the next generation's bit_n. This leads to a non-reversible cellular automaton since information about the previous generation is lost. Only 2^1 / 2^3 = 1/4 = 25% of the information is preserved from t-1 to t-0.
To make this non-reversible process into a reversible one, simply perform: bit_n at t-0 XOR bit_n at t-2. Even though it's only adding one more bit, it resolves the information loss problem. How? Shouldn't 50% of the information still be being lost generation to generation?
My guesses: either a reversible cellular automaton only loses 50% of the information from t-1 to t-0 instead of 75% (because the three t-1 bits that determine the next generation's bit_n are reused for other t-0 bits (for ex: bit_n-2, bit_n-1, bit_n at t-1 determine bit_n-1 at t-0)), and/or there is a correlation between bit_n at t-2 and bit_n at t-1.
Helpful link: https://en.wikipedia.org/wiki/Second-order_cellular_automaton#For_two-state_automata
