Question

construct a truth table for the given statement.
$(q \vee p) \wedge (\sim q \vee \sim p)$

complete the truth table.

p	q	~p	~q	$q \vee p$	$\sim q \vee \sim p$	$(q \vee p) \wedge (\sim q \vee \sim p)$
t	f	f	t	t	t	t
f	t	t	f	t	t	t
f	f	▼	▼	▼	▼	▼

Explanation:

Step1: Find ~p when p=F

When \( p = F \), the negation \( \sim p \) is \( T \) (since negation of false is true).

Step2: Find ~q when q=F

When \( q = F \), the negation \( \sim q \) is \( T \) (since negation of false is true).

Step3: Find \( q \vee p \) when p=F, q=F

The disjunction \( q \vee p \) is true if at least one of \( q \) or \( p \) is true. Since both \( p = F \) and \( q = F \), \( q \vee p = F \vee F = F \)? Wait, no: Wait, disjunction (OR) is true when at least one is true. Wait, \( F \vee F = F \)? Wait, no: Wait, \( T \vee F = T \), \( F \vee T = T \), \( T \vee T = T \), \( F \vee F = F \). Wait, but let's check again. Wait, the problem: when \( p = F \) and \( q = F \), \( q \vee p = F \vee F = F \)? Wait, but in the previous rows, when \( p = T, q = T \), \( q \vee p = T \); \( p = T, q = F \), \( q \vee p = T \); \( p = F, q = T \), \( q \vee p = T \). So for \( p = F, q = F \), \( q \vee p = F \vee F = F \).

Step4: Find \( \sim q \vee \sim p \) when p=F, q=F

\( \sim q = T \), \( \sim p = T \), so \( \sim q \vee \sim p = T \vee T = T \).

Step5: Find \( (q \vee p) \wedge (\sim q \vee \sim p) \) when p=F, q=F

We have \( q \vee p = F \) and \( \sim q \vee \sim p = T \). The conjunction (AND) is true only when both are true. So \( F \wedge T = F \). Wait, but let's re-express the original statement: \( (q \vee p) \wedge (\sim q \vee \sim p) \). Let's simplify the statement: \( (p \vee q) \wedge (\sim p \vee \sim q) \). This is equivalent to \( (p \wedge \sim q) \vee (q \wedge \sim p) \) (by distributive law: \( (p \vee q) \wedge (\sim p \vee \sim q) = p \wedge \sim p \vee p \wedge \sim q \vee q \wedge \sim p \vee q \wedge \sim q = F \vee (p \wedge \sim q) \vee (q \wedge \sim p) \vee F = (p \wedge \sim q) \vee (q \wedge \sim p) \)), which is the XOR (exclusive OR) of p and q. XOR is true when p and q have different truth values. So when p and q are both F (same), XOR is false. So let's verify:

For \( p = F, q = F \):

• \( \sim p = T \)
• \( \sim q = T \)
• \( q \vee p = F \vee F = F \)
• \( \sim q \vee \sim p = T \vee T = T \)
• \( (q \vee p) \wedge (\sim q \vee \sim p) = F \wedge T = F \)

Now let's fill the table:

Row 4 (p=F, q=F):

• ~p: T
• ~q: T
• \( q \vee p \): F
• \( \sim q \vee \sim p \): T
• \( (q \vee p) \wedge (\sim q \vee \sim p) \): F

So the missing values are:

~p: T

~q: T

\( q \vee p \): F

\( \sim q \vee \sim p \): T

\( (q \vee p) \wedge (\sim q \vee \sim p) \): F

Let's check the last column:

For the last row (p=F, q=F):

\( (q \vee p) = F \), \( (\sim q \vee \sim p) = T \), so their conjunction is \( F \wedge T = F \).

Answer:

For the row where \( p = F \) and \( q = F \):

• \( \sim p \): \( T \)
• \( \sim q \): \( T \)
• \( q \vee p \): \( F \)
• \( \sim q \vee \sim p \): \( T \)
• \( (q \vee p) \wedge (\sim q \vee \sim p) \): \( F \)

So the filled table cells (from left to right in the last row) are: \( T \), \( T \), \( F \), \( T \), \( F \)