# Show that $(p \wedge q)\rightarrow (p \vee q)$ is a tautology.

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 $F$
• Rules for disjunction :
• $A_3$: The statement $p\vee q$ has the truth value $F$ whenever both $p$ and $q$ have the truth value $F$.
• $A_4$: The statement $p\vee q$ has the truth value $T$ whenever either $p$ or $q$ or both have the truth value $T$
• 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 $q$
From the last columns of the truth table it is evident that $(p\wedge q)\rightarrow (p\vee q)$ is a tautology.
$\begin{matrix} p & q & p \wedge q & p \vee q & p \wedge q \rightarrow p \vee q \\ T & T & T& T & T \\ T & F & F& T & T \\ F & T & F & F & T\\ F & F & F & F & T \end{matrix}$
edited Mar 20, 2014