show in a truth table that p↔q and (p^q) v (¬p^¬q) are logically equivalent.
Truth table for (p∧q)∨(pˉ∧qˉ)(p\wedge q) \vee (\bar p\wedge \bar q)(p∧q)∨(pˉ∧qˉ) and p↔qp\leftrightarrow qp↔q:
∣pqp∧qpˉ∧qˉ(p∧q)∨(pˉ∧qˉ)p↔q000111010000100000111011∣\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}∣∣p0011q0101p∧q0001pˉ∧qˉ1000(p∧q)∨(pˉ∧qˉ)1001p↔q1001∣∣
Since these expressions takes equal values for all p,qp,qp,q, they are logically equivalent.
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