Using two column proofs prove sub4 and sub8 using only previ

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:

Solution

Sub4 : A-(B-A) = A

proof: A-(B-A)

=> A-B+A (removing braces)

=> A-B ( A+A = A)

=> A B^c    (Given as A-B = AB^c )

=> 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 B^c)

=> A (B1^c B2^c )       ((A U B)^c = A^c B^c )

=> (A B1^c ) (A B2^c ) (splitting brackets)

=> (A - B1) (A - B2)      (Given A - B = A B^c )

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 only