I am just wondering if its fair to say that A XNOR B is logically equivalent to A XOR ~B? If yes, how do I justify that without a truth table?

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    \$\begingroup\$ You can use boolean logic, expand those in terms of AND, OR and NOT operations and use De Morgan's laws and other logic rules to show they are equivalent, if they are equivalent. \$\endgroup\$ – jDAQ Apr 1 at 0:28
  • \$\begingroup\$ Thats the problem I'm having right now. I'm trying to verify that using boolean logic but Im not able to. I have verified the equivalence using truth tables and they are equivalent. I just dont know how to use boolean logic to do it. \$\endgroup\$ – AlwaysConfused Apr 1 at 0:33

A XNOR B: $$ \overline{A \oplus B} = \overline {A\cdot \overline B + \overline A \cdot B}$$

A XOR ~B: $$ A \oplus \overline B = A\cdot B + \overline A \cdot \overline B$$

Starting from A XNOR B, and using De Morgan's law: $$ \overline{A \oplus B} = \overline {A\cdot \overline B + \overline A \cdot B} \overset{DM}{=} {(\overline A + B) \cdot ( A + \overline B)} = { ( (\overline A + B) \cdot A + (\overline A + B) \cdot \overline B)}$$ $$ \overline{A \oplus B} = { ( (\overline A \cdot A+ B \cdot A) + (\overline A \cdot \overline B+ B \cdot \overline B))} = (\overline A \cdot A+ B \cdot A) + (\overline A \cdot \overline B+ B \cdot \overline B) $$

Notice some of those terms are contradictions, like \$\overline A \cdot A=0\$,se we can replace them $$ \overline{A \oplus B} = 0 +B \cdot A + \overline A \cdot \overline B +0.$$

With similar manipulations you should be able to get to the final answer, and even show the reverse (even though the steps are bidirectional).

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    \$\begingroup\$ Thank you, that really helps. That makes a lot of sense. I really appreciate the help. \$\endgroup\$ – AlwaysConfused Apr 1 at 0:55

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