In reply to rkg_:
( (o_fll_sar_eoc == 1) && i_fll_unlock_prog == 2'b01 ) |->
##1 (($rose(o_fll_inc)[=2] or $rose(fll_dec)[=2]) intersect 1'b1[*1:8])
##0 o_fll_lock;
// Rearranging the ()s
antecedent |->
##1 ( (a_sequence or b_sequence) intersect 1'b1[*1:8]
)
##0 o_fll_lock;
// Thus, in
($rose(o_fll_inc)[=2] or $rose(fll_dec)[=2]) intersect 1'b1[*1:8]
// the following two threads start at the same cycle
($rose(o_fll_inc)[=2] or $rose(fll_dec)[=2]) // One thread
1'b1[*1:8] // a parallel thread
intersect 1’b1[*1:8] , will it start matching just after sar_eoc==1. is it correct understanding ?
NO, sar_eoc==1 is in the antecedent. As mentioned above, the intersect starts after the ##1 at
($rose(o_fll_inc)[=2] or $rose(fll_dec)[=2])
Ben