I'm confused on the Logic Primer Files Exercise 3-2.d's answer. The answer manual says that we go from ~[(A&~B)v(C&~B)] to ~(A&~B)&~(C&~B) (De Morgan's law(s)) to (~AvB)&(~CvB) (De Morgan's law(s)) and substitution of logical equivalents) to (~AvB)&(~CvB) (double negation and substitution of logical equivalents) to (~A&~C)vB (distributive law and redundancy law). I got up to the second logic sentence in the list, but I don't see how De Morgan's law(s) apply to give the third. De Morgan's laws are ~(XvY) is logically equivalent to ~X&~Y and ~(X&Y) is logically equivalent to ~Xv~Y. The second logic sentence is of the form ~X&~Y, so it is logically equivalent to ~(XvY). But this is the first sentence's form, so applying it would simply send us back to the first sentence.

Even if this is explainable, I believe I have a proof which contradicts the answer. I got from ~[(A&~B)v(C&~B)] to ~(A&~B)&~(C&~B) (via De Morgan's law(s), and this is where I and the answer manual agree) to (~A&B)&(~C&B) (by distributing the negative) to (~A&B)&(~C&B) (via double negation and the substitution of logical equivalents) to ~A&B&~C&B (via the associative law), which will not contain an 'or' in it. Please help me!