How to prove this tautology using only logical equivalences? I'm having trouble with this problem.


  Using logical equivalencies prove that $(p \land q)\implies (p \lor q)$ is a tautology.
According to the equivalence $(a\rightarrow b) \Longleftrightarrow \neg a\vee b$, the De Morgan's laws, the associativity and commutativity of the logical operator $\vee$, one has

\begin{align*}
(p\wedge q) \longrightarrow (p\vee q) &amp; \Longleftrightarrow \neg(p\wedge q)\vee(p\vee q) \Longleftrightarrow (\neg p\vee \neg q)\vee(p\vee q)\\\\
&amp; \Longleftrightarrow (\neg p\vee p)\vee(\neg q\vee q) \Longleftrightarrow t\vee t \Longleftrightarrow t
\end{align*}

where $t$ represents a tautology.
A tautology is a statement that is always 'True'.

In your case you have two variables, each may take the value of either 'true' or 'false'.

You may simply plug these four cases in your expression and get 'True' in each case you try. Note that your logical expression is a composition of three operations, i.e. 'and', 'or' and 'implies'. Please, check with Wikipedia for their truth tables.

https://en.wikipedia.org/wiki/Modus_ponens