**Question:**(20 pts) Use the rules of inference to show that if ∀x(A(x) → B(x)), ∃x(¬B(x) ∨ C(x)) and ∀x ¬(D(x) ∨ ¬A(x)) are true, then ∃x(C(x) ∧ ¬D(x)) is also true, where the domains of all the quantifiers are the same.---**Solution Explanation:**To solve this problem, we'll use several rules of inference within the scope of predicate logic to derive the conclusion. Here are the steps broken down:1. **Given Statements:** - ∀x(A(x) → B(x)) - For all x, if A(x) is true, then B(x) is true. - ∃x(¬B(x) ∨ C(x)) - There exists an x such that ¬B(x) or C(x) is true. - ∀x ¬(D(x) ∨ ¬A(x)) - For all x, it is not the case that D(x) or ¬A(x) is true. Simplified, this means A(x) ∧ ¬D(x) is true for all x.2. **Objective:** - Show that ∃x(C(x) ∧ ¬D(x)) is true. - There exists an x such that both C(x) and ¬D(x) are true.3. **Proof Steps:** - From ∀x(A(x) → B(x)), assume A(x) is true, then B(x) must be true for all x. - ∃x(¬B(x) ∨ C(x)) implies that there exists some x for which ¬B(x) or C(x) holds: - If ¬B(x) is true, C(x) would need to be true for this x to satisfy ∃x(¬B(x) ∨ C(x)). - From ∀x ¬(D(x) ∨ ¬A(x)), we derive that A(x) ∧ ¬D(x) is true for all x. - This gives us that ¬D(x) is true wherever A(x) is true. - Combine these: - We find an x such that A(x) is true and

To prove that if , , and are true, then is also true, you can use rules of inference, specifically universal instantiation, existential instantiation, and modus ponens. Here's a step-by-step proof:1. Start with the given premises: a) b) c)2. Use existential instantiation (EI) on premise (b) to introduce a new constant symbol from the domain and derive: 3. Use universal instantiation (UI) on premise (a) to derive: 4. Use universal instantiation (UI) on premise (c) to derive: 5. Apply De Morgan's law to the expression in step 4: 6. Now, you have from step 2, and from step 3. Use modus ponens to derive: and imply7. Combine the result from step 6 with from step 5 to obtain: and imply8. Finally, use existential generalization (EG) to generalize the statement to the entire domain, which yields the desired conclusion: This proves that if the given premises are true, then is also true.Proof is given in explanation section