$(p \land q)\implies(p \lor q)$, how to make a truth table with $p$ twice? I thought I understood truth tables, until I saw repeats of $p$.  I guessed on this one: $p \implies \neg p$.  This is what I have:
$$ 
\begin{array}{c||c||c}
p & \neg p & p \implies \neg p\\
\hline
T & F & T\\
\hline
F & T & T
\end{array} 
$$
Now I am trying to write the truth table of  $(p \land q)\implies(p \lor q)$.  Can someone help me?
 A: Break your formula into smaller parts. In your example evaluate first $(p \land q)$ and $(p \lor q)$. Since you have two variables, the truth table has $2^2=4$ entries, on for each variable assignment. Filling out the table is now easy, proceed column by column.
$$ \begin{bmatrix}
p & q & (p\land q) & (p\lor q) & (p\land q) \implies (p\lor q) \\
F & F & F& F& T \\
F & T &  F & T& T\\
T & F & F&T& T\\
T & T & T &T& T\\
\end{bmatrix} $$
A: Well the first thing to do is to simplify the expression.
$(p \wedge q) \rightarrow (p \vee q)$
By material implication
$\neg (p \wedge q) \vee (p \vee q)$
By De Morgan's law
$(\neg p) \vee (\neg q) \vee (p \vee q)$
By associativity 
$((\neg p) \vee p) \vee ((\neg q) \vee q)$
Which is always true because $((\neg p) \vee p)$ is always true and $((\neg q) \vee q)$ is also always true.
So the logic table contains all T regardless of what p and q are assigned to.
Alternatively, you can painfully substitute, which I will start here:
$p = T$, $q = T$ gives:
$(T \wedge T) \rightarrow (T \vee T)$
$(T \wedge T) = T$, $(T \vee T) = T$, and $T \rightarrow T = T$
so this case is T.
You can do likewise for the other 3 cases.
A: When building the truth table for a formula, across each row you use the same truth values for the sentence variables that are used.  Think of this like evaluating a polynomial at a specific number.  If $$f(x) = 3x^2 - 5x + 6,$$ then to evaluate $f(2)$ we replace each instance of $x$ in the expression above by $2$, and then calculate as usual: $$f(2) = 3 \cdot 2^2 - 5 \cdot 2 + 6 = 12 - 10 + 6 = 8.$$
We do the same for each row of a truth table.  So let's evaluate some rows of the truth table of the formula you are interested in:
$$
\begin{array}{cc||l}
p & q & ( p \wedge q ) \Rightarrow ( p \vee q ) \\
\hline
T & F & ( T \wedge F ) \Rightarrow ( T \vee F ) = F \Rightarrow T = T \\
F & F & ( F \wedge F ) \Rightarrow ( F \vee F ) = F \Rightarrow F = T
\end{array}
$$
