Answer: a) FALSE b) FALSE
Step-by-step explanation:
a) For the given proposition p ∧ ¬(q ∨ s) you can solve first (q v s)
q v s is true if either q is <em>true</em> or s is <em>true </em>or both. It is only <em>false</em> if both q and s are <em>false</em>. So, the proposition (q v s) is <em>true</em> because q is <em>true</em>.
Now you can solve the negation: ¬(q ∨ s)
As we know, (q v s) is <em>true </em>then its negation ¬(q ∨ s) is <em>false</em>.
p ∧ ¬(q ∨ s) should be<em> true </em>when both p and ¬(q ∨ s) are <em>true</em>, and <em>false </em>otherwise. So, the proposition is<em> false</em> because p is <em>true</em> and ¬(q ∨ s) is <em>false</em>.
b) For the given proposition ¬(q ∧ p ∧ ¬s)
You can rewrite the expression as: ¬[q ∧ (p ∧ ¬s) ] to solve first each part of the propositions in parenthesis.
The negation of s: ¬s is <em>true</em> because s is <em>false</em>
Now, you can solve (p ∧ ¬s) which is <em>true</em> because both p and ¬s are<em> true.</em>
To continue, you have to solve (q ∧ p ∧ ¬s) which is<em> true</em> because both q and (p ∧ ¬s) are <em>true.</em>
To finish, the negation: ¬(q ∧ p ∧ ¬s) is <em>false</em>.