C: $K_A m; \\ K_B d.$
A: $\neg K_A d; \\ m \vDash \neg K_B m.$
B: $d \not\vDash K_B m; \\ (K_A(\neg K_B m)) \vDash K_B (m,d).$
A: $m \wedge K_B(m,d) \vDash K_A (m,d).$
Albert, Bernard and Cheryl have had a busy week. They’re the stars of #thatlogicproblem, a question from a Singapore maths test that was posted to Facebook by a TV presenter and quickly sent the internet deduction-crazy.
First of all: no, it’s not meant to be answered by an average Singaporean student. It’s a hard question from a schools Olympiad test.