I'm looking to use laws of set algebra to simplify:
$(A-B^c)∪(B∩(A∩B)^c)$
But I'm currently stuck on what I should do.
= $(A-B^c)∪(B∩(A^c∪B^c))$ (De Morgan's)
= $(A-B^c)∪((B∩A^c)∪(B∩B^c))$ (Distributive Law)
= $(A-B^c)∪(B∩A^c)$ (Intersection with Compliment)
And from here, I'm stuck. If I draw up a Venn Diagram, I can see that the simplified expression would just be $B$, but I'm currently stuck on how to get there. I've tried turning $(B∩A^c)$ to $B - A$, but that didn't seem to help me much.
On
$(A−B^c)∪(B∩(A∩B)^c)$
First go with De-Morgan's:
$B ∩ (A ∩ B)^c$ = $B ∩ (A^c ∪ B^c)$
Applying Distributive Law, $B ∩ (A^c ∪ B^c)$ = $(A^c ∩ B) ∪ (B ∩ B^c)$ = $(A^c ∩ B)$ - (1)
As, $A - B$ = $A ∩ B^c$. Therfore, $A - B^c$ = $A ∩ B$ -(2)
Using (1)&(2),
$(A ∩ B) ∪ (A^c ∩ B)$ = $(A ∩ A^c) ∪ B$ = $B$
We have $$(A-B^c)\cup(B\cap A^c)=(A\cap B)\cup(B\,\,\text\ A)=B$$