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# Show that p ↔ q ≡ (p → q) ∧ (q → p)

Toolbox:
• If $p$ and $q$ are two simple statements $p\wedge$ is the conjunction of $p$ and $q$ and $p \vee q$ is the disjunction .Negation of a statement $p$ is denoted by $\sim p$
• Rules for conjunction :
• $A_1$: The statement $p\wedge q$ has the truth table value $T$ whenever both $p$ and $q$ have the truth value $T$
• $A_2$: The statement $p\wedge q$ has the truth value $F$ whenever either $p$ or $q$ or both have the truth value $F$
• Conditional statement :"If $p$ then $q$" is written as $p\rightarrow q$ (p implies q).$p\rightarrow q$ is false only if $p$ is true and $q$ is false.If $p$ is false,then $p\rightarrow q$ is true ,regardless of the truth value of $q$
• Bi-conditional statement : If $p$ and $q$ are two statements then the compound statement $p\rightarrow q$ and $q\rightarrow p$ is a bi-conditional statement,written as $p \leftrightarrow q$ (p if and only of q).$p\rightarrow q$ has the truth value $T$ whenever $p$ and $q$ have the same truth values,otherwise it is $F$
The last two columns of the truth table are identical.
$\therefore p\rightarrow \equiv (p\rightarrow q\wedge (q\rightarrow p)$