I12-1 FR' F'R UF' U'F RU' R'U I12-2 FR' F'R UF' F'L FL' U'F I12-3 FR' F'R UF' UL' U'L FU'
I can't believe I'm the first person to notice this:
Suppose we only know I12-1 and I12-2. Then we have I12-1' U'RUR'F'U (FR'F'RUF')' (I12-1')(I12-2) U'RUR'F'U (FR'F'RUF')' (FR'F'RUF') F'LFL'U'F Reduce: U'RUR'F'U F'LFL'U'F Conjugating by (U'RUR'F'U)', we get F'LFL' U'FU'R UR'F'U But this is just the RL mirror image of FR'F'L UF'UL' U'LFU'
This is exactly I12-3. So there are really only two independent 12q
identities, and the third can be deduced from them.