Using two column proofs prove sub4 and sub8 using only previous results and axio
ID: 3406443 • Letter: U
Question
Using two column proofs prove sub4 and sub8 using only previous results and axioms.
And its relation to the other operations: Ctm6. ec A and A cU. Boundary relations Ctm7. A C B Bc Ac. Complements reverse containment Ctm18. ACB (A UC) C (BUC)and AC B (A n C) C (BnC). Ctm9. (Ai C B, and A2 c B) (As uA2jc (Bi U B2) and Derived Operations: Difference and Sum There is a derived binary operation A-B (set subtract) that is related to the complement. We may define set for set subtract are as follows: subtract by A-B An Ba. Some identities Subl. AC-U-A Complement in terms of subtract. Sub2, A-(A-B) An B. Intersection in terms of subtract. Sub3. (Ac-B) AUB. Union in terms of subtract. Sub4. A- (B- A) A. Sub5. AC-Bo B-A. Sub7. (Ai u A2)-B = (Al-B) U (12-B) and Sub8. A- (B UBa) (A-B)n (A- B2) and 4-(B, n Ba) = (A-BJU (A-Ba). DeMorgan Laws. There is a derived binary operation A+ B (set sum) that is used in making a ring out of this system using set sum as the addition and intersection as the multiplication. We may define set sum by A + B = (A-B) U (B-A). There is no need for a symbol -A because every set is its own additive inverse. Some identities for set sum are as follows:Explanation / Answer
Sub4 : A-(B-A) = A
proof: A-(B-A)
=> A-B+A (removing braces)
=> A-B ( A+A = A)
=> A Bc (Given as A-B = ABc )
=> A
Two-Column proof:
sub 8: Demorgan laws
i) A - (B1 U B2) = (A - B1) (A - B2)
proof:
=> A - (B1 U B2)
=> A (B1 U B2)c (Given A - B = A Bc)
=> A (B1c B2c ) ((A U B)c = Ac Bc )
=> (A B1c ) (A B2c ) (splitting brackets)
=> (A - B1) (A - B2) (Given A - B = A Bc )
2 column proof:
ii) A - (B1 B2) = (A - B1) U (A - B2)
2 lines proof:
Statement Theorem and reasons A - (B - A) Statement Given A - B + A Removing braces and multiplying minus to inside elements of brace A - B A + A = A A Bc Given that A - B = A Bc A A and Bc has common area which is A onlyRelated Questions
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