three_SIG0_nonconsecutive = {SIG0[=3]}

This sequence looks for 3 nonconsecutive occurrences of SIG0 since we used [=n] instead of the consecutive operator [*n]. Note that SIG0 does not have to be true during the first cycle of the match attempt, this operator only that we eventually see 3 clocks during which SIG0 is true, so we end up with a large number of successful matches.