For set problems like this, it helps if you can use a Venn diagram. In this case, B(union)Ac would be everything shaded outside of A and B, plus everything in B (including the intersection of A and B). Using this:
Bc(union)A is not equivalent because it includes all of A.
B(union)[(Bc(intersect)Ac)] is equivalent.
(A(intersect)Bc)c is equivalent.
Bc(intersect)A is not equivalent because it's actually the complement of the original statement.
Ac(union)B is equivalent because it's the same as the original statement, just in reverse order, and set expressions are commutative.
Sorry for the cumbersome notation, but I'm not able to write symbols.
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For set problems like this, it helps if you can use a Venn diagram. In this case, B(union)Ac would be everything shaded outside of A and B, plus everything in B (including the intersection of A and B). Using this:
Bc(union)A is not equivalent because it includes all of A.
B(union)[(Bc(intersect)Ac)] is equivalent.
(A(intersect)Bc)c is equivalent.
Bc(intersect)A is not equivalent because it's actually the complement of the original statement.
Ac(union)B is equivalent because it's the same as the original statement, just in reverse order, and set expressions are commutative.
Sorry for the cumbersome notation, but I'm not able to write symbols.