Consider the following proof that for any sets A, B [A-B=0] Proof. [AC B] (0) We'll...

  

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Consider the following proof that for any sets A, B [A-B=0] Proof. [AC B] (0) We'll prove the equivalent contrapositive, namely that [A-B0] [AB] (1) Suppose A- B+0 (2) there exists an element a A - B (3) a A and a B (4) it is not the case that if b = A then b B (5) it is not the case that ACB (6) AB For each of the steps below, match it to the reason that explains why that step is valid. 1. by substitution 2. by algebra 3. by def of mod 4. by definition of multiple / divides 5. because the difference and product of integers is integer (1)-(2) (2)-(3) (4)-(5) (5)-(6) 6. by the definition of n 7. by the definition of U 8. by the definition of E 9. by the definition of \(set minus) (also written -) 10. by the definition of set complement 11. by definition of 12. by definition of 13. by def of 14. because the negation of "there is not" is "there is" (and vice-versa) Consider the following proof that for any sets A, B [A-B=0] Proof. [AC B] (0) We'll prove the equivalent contrapositive, namely that [A-B0] [AB] (1) Suppose A- B+0 (2) there exists an element a A - B (3) a A and a B (4) it is not the case that if b = A then b B (5) it is not the case that ACB (6) AB For each of the steps below, match it to the reason that explains why that step is valid. 1. by substitution 2. by algebra 3. by def of mod 4. by definition of multiple / divides 5. because the difference and product of integers is integer (1)-(2) (2)-(3) (4)-(5) (5)-(6) 6. by the definition of n 7. by the definition of U 8. by the definition of E 9. by the definition of \(set minus) (also written -) 10. by the definition of set complement 11. by definition of 12. by definition of 13. by def of 14. because the negation of "there is not" is "there is" (and vice-versa)