Question

show in a truth table that p↔q and (p^q) v (¬p^¬q) are logically equivalent.

Answer 1

Truth table for $(p\wedge q) \vee (\bar p\wedge \bar q)$ and $p\leftrightarrow q$:

$\begin{vmatrix} p & q & p\wedge q & \bar p\wedge \bar q & (p\wedge q) \vee (\bar p\wedge \bar q) & p\leftrightarrow q\\ 0 & 0 & 0 & 1 & 1 & 1\\ 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 0 & 1 & 1\\ \end{vmatrix}$

Since these expressions takes equal values for all $p,q$, they are logically equivalent.