Question

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

p	q	$sim q$	$sim q wedge p$	$q wedge p$	$(sim q wedge p) vee (q wedge p)$
t	f	$\blacktriangledown$	$\blacktriangledown$	$\blacktriangledown$	$\blacktriangledown$

Explanation:

Step1: Find ~q when p=T, q=F

Negation of q (F) is T, so ~q = T.

Step2: Calculate ~q ∧ p when ~q=T, p=T

Conjunction (∧) is T when both are T, so ~q ∧ p = T ∧ T = T.

Step3: Calculate q ∧ p when q=F, p=T

Conjunction (∧) is F when one is F, so q ∧ p = F ∧ T = F.

Step4: Calculate (~q ∧ p) ∨ (q ∧ p) when ~q ∧ p=T, q ∧ p=F

Disjunction (∨) is T when at least one is T, so T ∨ F = T.

Answer:

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

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

So the completed row (from left to right in the empty cells) is \( T \), \( T \), \( F \), \( T \).