# $(p \land q)\implies(p \lor q)$, 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?

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For some basic information about writing math at this site see e.g. here, here, here and here. I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. –  Julian Kuelshammer Jan 11 '13 at 8:07
Break it up: First the two usual columns: $p$, $q$. Then build the $p \land q$, and the $p \lor q$ columns from the values in the first two columns. Then build the final column based on columns 3 and 4. –  Peter Grill Jan 11 '13 at 8:16

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.

$$\begin{bmatrix} p & q & (p\land q) & (p\lor q) & (p\land q) \implies (p\lor q) \\ F & F & F& F& T \\ F & T & F & T& T\\ T & F & F&T& T\\ T & T & T &T& T\\ \end{bmatrix}$$

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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.

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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: $$\begin{array}{cc||l} p & q & ( p \wedge q ) \Rightarrow ( p \vee q ) \\ \hline T & F & ( T \wedge F ) \Rightarrow ( T \vee F ) = F \Rightarrow T = T \\ F & F & ( F \wedge F ) \Rightarrow ( F \vee F ) = F \Rightarrow F = T \end{array}$$

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