A valid argument is given below. Fill in the missing values of the proof. vx (R(x) v Q(x)) 3X (P(x) v -Q(x)) 1. ax (P(x) v -Q(x)) Hypothesis 2. (b is a particular element) ^ (P(b) v ¬Q(b)) (A) : 3X (P(x) v R(x)) 3. P(b) v -Q(b) (B) 4. vx (R(x) v Q(x)) Hypothesis 5. b is a particular element (C) 6. (D) Universal instantiation, 4, 5 7. P(b) v R(b) Resolution, 3, 6 8. 3X (P(x) v R(x)) Existential generalization, 5, 7 (A) Pick (C) [Pick (B) Pick (D) Pick

Q: laws of logic to show that ((s VrV¬r) ^ (SV¬rva) ^ (rv¬sva)) = q v¬s.

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A: Given two logic expressions are, P→Q, (R→P)→(R→Q) P→Q can be written as ¬P∨Q

Q: - f(n) = log n²; g(n) = log n + 5 - f(n) = n; g(n) = log n? - f(n) = log log n; g(n) = log n f(n) =…

A: analysis for the above example is in step 2.

Q: A valid argument is given below. Fill in the missing values of the proof. Vx(P(x)→ Q(x)) 3x…

A: Here, a proof applying the predicate logic is given. The task is to fill the missing values to the…

Q: ¬[[p → ¬q) v s] = [p Aq A -s] -[[p → ¬q) → s] = [¬(p^q) A¬s] [(pA¬q) → s] = [¬s → (-p v q)] [¬s →…

A:

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A: pq~p~qp↔q∼q→∼pq→∼p∼q→∼p & q→∼pTTFFTTTTTFFTFFTFFTTFFTFFFFTTTTFTclearly option A is correct.

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Q: Prove by natural deduction: Show ∃x[p(x) ∧ q(x)] ⊧ ∃xp(x) ∧ ∃xq(x)

A: Prove by natural deduction: Show ∃x[p(x) ∧ q(x)] ⊧ ∃xp(x) ∧ ∃xq(x)
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Q: Write a logical proof to prove that the following argument is valid. Vx(P(x) → Q(x)) 3x(R(x) ^ P(x))…

A: Uploading handwritten answer