Questions: Complete the following truth table. Use T for true and F for false. You may add more columns, but those added columns will not be graded. p q ~p ↔ q q → ~p ---------------------- T T T F F T F F

Complete the following truth table. Use T for true and F for false.
You may add more columns, but those added columns will not be graded.

p q ~p ↔ q q → ~p
----------------------
T T
T F
F T
F F

Transcript text: Complete the following truth table. Use T for true and F for false. You may add more columns, but those added columns will not be graded. \begin{tabular}{|c|c|c|c|} \hline$p$ & $q$ & $\sim p \leftrightarrow q$ & $q \rightarrow \sim p$ \\ \hline T & T & $\square$ & $\square$ \\ \hline T & F & $\square$ & $\square$ \\ \hline F & T & $\square$ & $\square$ \\ \hline F & F & $\square$ & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

To complete the truth table, we need to evaluate the logical expressions for each combination of truth values for $ p $ and $ q $. Specifically, we need to determine the values of $ \sim p \leftrightarrow q $ and $ q \rightarrow \sim p $ for each row in the table.

For each combination of $ p $ and $ q $:
- Calculate $ \sim p $ (the negation of $ p $).
- Evaluate $ \sim p \leftrightarrow q $ (biconditional: true if both sides are the same).
- Evaluate $ q \rightarrow \sim p $ (implication: false only if $ q $ is true and $ \sim p $ is false).

Step 1: Define the Truth Values

We consider the truth values for $ p $ and $ q $ as follows:

$ p = T $, $ q = T $
$ p = T $, $ q = F $
$ p = F $, $ q = T $
$ p = F $, $ q = F $

Step 2: Calculate $ \sim p $

The negation of $ p $ is calculated for each case:

If $ p = T $, then $ \sim p = F $.
If $ p = F $, then $ \sim p = T $.

Step 3: Evaluate $ \sim p \leftrightarrow q $

We evaluate the biconditional $ \sim p \leftrightarrow q $ for each combination:

For $ (T, T) $: $ \sim p = F $ and $ q = T $ gives $ F \leftrightarrow T = F $.
For $ (T, F) $: $ \sim p = F $ and $ q = F $ gives $ F \leftrightarrow F = T $.
For $ (F, T) $: $ \sim p = T $ and $ q = T $ gives $ T \leftrightarrow T = T $.
For $ (F, F) $: $ \sim p = T $ and $ q = F $ gives $ T \leftrightarrow F = F $.

Step 4: Evaluate $ q \rightarrow \sim p $

Next, we evaluate the implication $ q \rightarrow \sim p $:

For $ (T, T) $: $ q = T $ and $ \sim p = F $ gives $ T \rightarrow F = F $.
For $ (T, F) $: $ q = F $ and $ \sim p = F $ gives $ F \rightarrow F = T $.
For $ (F, T) $: $ q = T $ and $ \sim p = T $ gives $ T \rightarrow T = T $.
For $ (F, F) $: $ q = F $ and $ \sim p = T $ gives $ F \rightarrow T = T $.

Step 5: Compile the Results

The completed truth table is as follows:

\[ \begin{array}{|c|c|c|c|} \hline p & q & \sim p \leftrightarrow q & q \rightarrow \sim p \\ \hline T & T & F & F \\ T & F & T & T \\ F & T & T & T \\ F & F & F & T \\ \hline \end{array} \]

Final Answer

The completed truth table values are:

For $ (T, T) $: $ \sim p \leftrightarrow q = F $, $ q \rightarrow \sim p = F $
For $ (T, F) $: $ \sim p \leftrightarrow q = T $, $ q \rightarrow \sim p = T $
For $ (F, T) $: $ \sim p \leftrightarrow q = T $, $ q \rightarrow \sim p = T $
For $ (F, F) $: $ \sim p \leftrightarrow q = F $, $ q \rightarrow \sim p = T $

Thus, the final answer is: \[ \boxed{ \begin{array}{|c|c|c|c|} \hline p & q & \sim p \leftrightarrow q & q \rightarrow \sim p \\ \hline T & T & F & F \\ T & F & T & T \\ F & T & T & T \\ F & F & F & T \\ \hline \end{array} } \]

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