11. (10 pts) the following is an outline of a proof that (a n by ca u b. fill in the blanks. proof: given sets a and b, to prove that (an by s a ub we suppose xea c and then we show that x e so suppose that then by definition of complement, .so by definition of intersection, it is not the case that (x is in a and x is in b). consequently, x is not in b because of de morgan's law of logic. x is not in a in symbols, this says that x ea or x eb. again, by definition of complement, x e thus, by definition of union, x e [as was to be shown]. theorem 6.2.2. set identities (b) anb= bna (b) (an bn c= an (bn c (b) an (buc) (an bu (anc) (a) aub=bua (a) (a u buc au(bu c 1. commutative laws: 2. associative laws: 3. distributive laws: 4. identity laws: (a) au (bnc) ( aub)n(auc) (a) au0= a (a) au a u (a a (a) au a a (a) auu u (a) (a u b anb (a) au (an b = a (a) u 0 a b a nb a (b) anu (b) an a 0 5. complement laws: 6.double complement law 7. idempotent laws: 8. universal bound laws: (b) an a=a (b) ane 0 (b) (an baub (b) an (a u b = a (b) u 9. de morgan's laws: 10. absorption laws: 11.complements of u and 0 12. set difference law: 12. (10 pts) derive the following result "algebraically" using the properties listed in theorem 6.2.2, give a reason for every step that exactly justifies what was done in the step. "for all sets a, b, and c, (a u c) - b (a- b) u (c-b)."