1. tL2 < tL3 (on process i), and
2. te < tw < tc (on process k),
(in the non-trivial case) tc can only occur before tL2 (tc < tL2) => tw < tL3, that is process i reads the current value of number[k]. See, no overlap! "The bakery algorithm is correct as long as reading a number returns the correct value if the number is not concurrently being written".