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## Comments

Let \(a,b \in P\) and subset \(A = \{ a \leq b \; and \; b \leq a \}\). Then you get two meets namely, \(a \wedge b = a\) and \(b \wedge a = b\)?

Using reflexivity and transitivity, we can expand this out into a transitivity triangle and get 1. \(a \leq b \leq a\) and \(a \leq a\) and 2. \(b \leq a \leq b\) and \(b \leq b\). Taking \(a \wedge b\) in Triangle 1 and Triangle 2 gives a and b respectively.

`Let \\(a,b \in P\\) and subset \\(A = \\{ a \leq b \; and \; b \leq a \\}\\). Then you get two meets namely, \\(a \wedge b = a\\) and \\(b \wedge a = b\\)? Using reflexivity and transitivity, we can expand this out into a transitivity triangle and get 1. \\(a \leq b \leq a\\) and \\(a \leq a\\) and 2. \\(b \leq a \leq b\\) and \\(b \leq b\\). Taking \\(a \wedge b\\) in Triangle 1 and Triangle 2 gives a and b respectively.`