Chapter 6 - Set Theory - Exercise Set 6.3 - Page 373: 32

See explanation

We have to prove: $For\,\,all\,\,sets\,\,A,\,B,\,and\,C,\\$ $$(A - B) \cup (A \cap B) = A.$$ Proof: $(A - B) \cup (A \cap B)=(A \cap B^c) \cup (A \cap B)\\ by\,\,\,(set\,\,dif\! ference\,\,law )\\ =A \cap (B^c \cup B)\\ by\,\,\,(distributive\,\,law)\\ \\ =A \cap U\\ by\,\,\,complement\,\,law \\ =A\\ by\ identity\,law\,for\,\cap )\\ $