Difficulty: Hard · Puzzle 5 of 6 · Topic: Sequence Puzzles
Property intent: both branches must match with the same start and same end using intersect. Timeline: C50 start=1 req=1, C51 gnt=1, C52 done=1. Determine the verdict.
property p_seq_intersect;
@(posedge clk) start |-> ((req ##1 gnt) intersect (req ##2 done));
endpropertyIntersect is stricter than and: same start and same end cycle.
1) Left branch ends at C51 (req ##1 gnt).
2) Right branch ends at C52 (req ##2 done).
3) End times differ by one cycle.
4) intersect therefore has no valid match.Answer: FAIL - both branches start together, but they end on different cycles, violating intersect semantics.
Candidates who memorize only logical intuition miss that intersect constrains temporal end alignment.
What minimal code change would make this puzzle pass while keeping intersect?