Forward direction
Take any \(p \in P\) and let \(q = f(p) \in Q\)
By reflexivity, we have in \(Q\) that \(f(p) \leq q\)
By definition of Galois connection, we then have \(p \leq g(q)\), so (1) holds.
Take any \(q \in Q\) and let \(p = g(q) \in P\)
By reflexivity, we have in \(P\) that \(p \leq g(q)\)
By definition of Galois connection, we then have \(f(p) \leq q\), so (2) holds.
Reverse direction
Want to show \(f(p)\leq q \iff p \leq g(q)\)
Suppose \(f(p) \leq q\)
Since g is monotonic, \(g(f(p)) \leq g(q)\)
but, because (1), \(p \leq g(f(p))\), therefore \(p \leq g(q)\)
Suppose \(p \leq g(q)\)
Since f is monotonic, \(f(p) \leq f(g(q))\)
but, because (2), \(f(g(q)) \leq q\), therefore \(f(p) \leq q\)