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Multiply both sides by pq u q u p = m p q u (qu) = m p q u = m p q / (qu) that would be undefined (bottom = 0) if p = q. Featured Products Pro Preferred Professional installers demand the BEST, so look for our QEP Pro Preferred line of professional products Learn More. If there is job 1 in P ways and job 2 in q ways and both are related, we can do only 1 job at given time in pq ways Multiplication Rule in Probability If there is job 1 in P ways and job 2 in q ways and both are not related, we do both jobs at given time in p*q ways Learn how the concepts of midpoint and arithmetic sequence differs from.

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Example 232 Show (p!q) is equivalent to p^q Solution 1 Build a truth table containing each of the statements p q q p!q (p!q) p^q T T F T F F T F T F T T F T F T F F F F T T F F Since the truth values for (p!q) and p^qare exactly the same for all possible combinations of truth values of pand q, the two propositions are equivalent. Q s b { ÿ. _ P M 5 2.

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_ / Æ. 11 PROPOSITIONS 7 p q ¬p p∧q p∨q p⊕q p → q p ↔ q T T F T T F T T T F F F T T F F F T T F T T T F F F T F F F T T Note that ∨ represents a nonexclusive or, ie, p∨ q is true when any of p, q is true and also when both are true On the other hand ⊕ represents an exclusive or, ie, p⊕ q is true only when exactly one of p and q is true 112. Feb 27, 16There's what I have so far We assume that P(A) ⊆ P(B) This means that every element x that exists in P(A), also exits in P(B) By definition of a power set, x∈P(A) if x ⊆ A Therefore, A∈P(A) Since P(A) ⊆ P(B), A∈P(B), meaning all x ⊆ A, x ∈ P(B) Furthermore, B∈P(B), meaning all x ⊆ B, x ∈.

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