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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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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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