Simplifying Boolean $ab+c+(abc)'+bc+c'b$ and $xy+xz+(yz+yz')'$

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I have these two equations $$ab +c + (abc)' + bc + c'b$$ $$xy + xz + (yz + yz')'$$ and need to simplify both. Addition means $\lor$, multiplication means $\land$ and the prime means $\neg$.

I have been reading that I have to follow certain rules but I can't get my head arount it. Can somebody help show me how to simplify both?

Thanks in advance.

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$$ab+c+(abc)'+bc+c'b$$ Use De Morgan's law on $(abc)'$: $$=ab+c+a'+b'+c'+bc+c'b$$ Since $x+x'=1$: $$=ab+1+a'+b'+bc+c'b$$ This is a disjunction of terms, one of which is a 1, so the whole thing immediately reduces to 1.


$$xy+xz+(yz+yz')'$$ Collect $y$ in $(yz+yz')'$, then use $x+x'=1$: $$=xy+xz+y'$$ Distribute $\lor$ over $\land$ in $xy+y'$, then use $x+x'=1$ (again): $$=x+y'+xz$$ $x$ absorbs $xz$: $$=x+y'$$ This is the most simplified form.