# Boolean Logical Algebra - Prove 4 nor gates to an xnor gate.

Need to reach the following conclusion (or maybe its the premise?)

AB + A'B' = F

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It is a truth table analysis. We have to show that the combination of NOR gates results in an XNOR gate. So we have to show that the combination returns $1$ precisely if $A$ and $B$ both have value $0$ or both have value $1$.
We need to use the definition of NOR gate. A NOR gate returns $1$ precisely if both inputs are $0$. In any other situation, a NOR gate returns $0$.
Start with $A=0$ and $B=0$. Then the first NOR gate returns $1$. But then the two next NOR gates (the ones on top of each other) each return $0$. and then the final gate returns $1$.
Do this analysis for all four possibilities for $A$ and $B$. Actually, you only need three, since there is symmetry between $A$ and $B$. So the only remaining cases are effectively $A=0$, $B=1$ and $A=1$, $B=1$.