True or false? Prove or justify, respectively. "Given $$f(x,y,z)=x\cdot z+\overline y\cdot\overline{z+\overline x}$$ can be implemented in a circuit formed only by an $\mathrm{AND}$ gate, an $\mathrm{OR}$ and a $\mathrm{NOT}$".
I think it is false.
\begin{align*} f(x,y,z)&=x\cdot z+\overline y\cdot\overline{z+\overline x}\\ &=x\cdot z+\overline y\cdot\overline z\cdot x\\ &=x\cdot z+\overline y\cdot\overline z\\ &=(x\cdot z)+(\overline y\cdot\overline z), \end{align*}
so the minimum expression of $f$ has two $\mathrm{AND}$s, one $\mathrm{OR}$ and two $\mathrm{NOT}$s.
Even if we try to apply De Morgan's Law to the original expression of $f$ we end up with $$\overline{x+\overline z\cdot y+(z+\overline x)}\implies\overline{x+\overline x+\overline z\cdot y+z}\implies\overline{\mathrm T+\overline z\cdot y+z}\implies\overline{\mathrm T}\implies\mathrm C.$$ Is any of the "answers" correct? How would you do it?
Thanks!