Circuit Analysis using Boolean Algebra:

• The top NAND gate has both of its inputs tied together to input $A$. Its output is:
  $$Y_1 = \overline{A \cdot A} = \bar{A}$$
• The bottom NAND gate has both of its inputs tied together to input $B$. Its output is:
  $$Y_2 = \overline{B \cdot B} = \bar{B}$$
• These two inverted outputs are fed into the third NAND gate. The final output $Y$ is:
  $$Y = \overline{Y_1 \cdot Y_2} = \overline{\bar{A} \cdot \bar{B}}$$
• Applying De Morgan's Law ($\overline{X \cdot Y} = \bar{X} + \bar{Y}$):
  $$Y = \overline{\bar{A}} + \overline{\bar{B}} = A + B$$
  This is precisely the Boolean expression for an OR gate.