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    \$\begingroup\$ Long time no see isaac! \$\endgroup\$ Commented Jul 11, 2018 at 1:36
  • \$\begingroup\$ When \$P, Q\$ are two subsets, is there any scenario where \$P \leq Q\$ can be deduced from any information other than \$P \subseteq Q\$ or \$\forall p \in P, q\in Q (p \leq q)\$, not counting the initial \$a \leq b \leq c \leq \dots \$? \$\endgroup\$ Commented Jul 11, 2018 at 2:14
  • \$\begingroup\$ Answer: yes. \$\forall p \in P, q\in Q (p \leq q)\$ is not tight enough, example: \$\{a, c\}, \{b, c\}\$. \$\endgroup\$ Commented Jul 11, 2018 at 2:22
  • \$\begingroup\$ @orlp Good to be back! I think I'll be doing mostly questions for the foreseeable future \$\endgroup\$ Commented Jul 11, 2018 at 3:47
  • \$\begingroup\$ Could you also add the 14 possible orderings for n=4? \$\endgroup\$ Commented Jul 11, 2018 at 12:35