**34**: there is no pair shared by both P and Q, so the meet is discrete.

**35**: conjunction of equivalence relations preserves reflexivity, symmetry, and transitivity (x~y~z P and x~y~z Q implies x~z P and x~z Q, etc.)

**36**: clearly this conjunction is finer than each. suppose there were another such relation R that was coarser; but then there exists x,y such that x~y P and x~y Q, but x!~y R, so it is not finer, contradiction. hence the conjunction is the coarsest relation which is finer than P and Q, thus the "meet" in the poset.

**37**: the problem is with transitivity - if x~y P and y~z Q, that does not imply x~z in either. so the naive disjunction of relation "breaks" along these points, and is not transitive.

**38**: so, if we add transitive closure, then this ignores the ~ pairs which span across the break points. we then have an equivalence relation, because reflexivity and symmetry were trivially satisfied in the definition.

**39**: clearly the disjunction is coarser than each, and the transitive closure only eliminates some of the new "joined" ~s which failed to be transitive. suppose there were another such relation R that was finer; then either transitivity is lost, or there exists x~y R but x!~y P and x!~ Q, so it is not coarser, contradiction. hence the transitive closure of the disjunction is the finest relation which is coarser than P and Q, thus the "join" in the poset.

someone please feel free to make these nicer! (or find mistakes)

**35**: conjunction of equivalence relations preserves reflexivity, symmetry, and transitivity (x~y~z P and x~y~z Q implies x~z P and x~z Q, etc.)

**36**: clearly this conjunction is finer than each. suppose there were another such relation R that was coarser; but then there exists x,y such that x~y P and x~y Q, but x!~y R, so it is not finer, contradiction. hence the conjunction is the coarsest relation which is finer than P and Q, thus the "meet" in the poset.

**37**: the problem is with transitivity - if x~y P and y~z Q, that does not imply x~z in either. so the naive disjunction of relation "breaks" along these points, and is not transitive.

**38**: so, if we add transitive closure, then this ignores the ~ pairs which span across the break points. we then have an equivalence relation, because reflexivity and symmetry were trivially satisfied in the definition.

**39**: clearly the disjunction is coarser than each, and the transitive closure only eliminates some of the new "joined" ~s which failed to be transitive. suppose there were another such relation R that was finer; then either transitivity is lost, or there exists x~y R but x!~y P and x!~ Q, so it is not coarser, contradiction. hence the transitive closure of the disjunction is the finest relation which is coarser than P and Q, thus the "join" in the poset.

someone please feel free to make these nicer! (or find mistakes)