Using either a truth table, or resolution, check if the following conclusion is valid. Given 1. A ⇒ (B V C) 2. A^ D 3. D⇒ ~C Conclusion • B Hint: resolution is the shorter approach.
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- Determine whether or not the following statement is a tautology or not and give reasoning. If you need to, you can build a truth table to answer this question. (q→p)∨(∼q→∼p) A. This is a tautology because it is always true for all truth values of p and q. B. This is not a tautology because it is always false for all truth values of p and q. C. This is a tautology because it is not always false for all values of p and q. D. This is not a tautology becasue it is not always true for all truth values of p and q.Will be made avallable the day after deadline Question 13 To solve this proof, what rules would you need to apply. Please type rule abbreviations (MP MT DS HS CD Conj Simp Add DM Dist Assoc) in the order in which you applied them (from left to right, leaving a space between rules). Example: MT Simp Add MP 1. (N Z) v (N •~P) 2. (ZƆ S) • (~P Ɔ E) /SVE 44 F7 * FS de * F2 F3 * & % 8. 2$Direction: Determine whether the given statement is a tautology, contradiction, or contingency. Construct a truth table. p →~p (p Λ q) →p (p → q) Λ (q → p) (p →~q) Λ (~p →q)
- 6. Use a truth table to determine whether the following argument form is valid. Indicate which columns represent the premises and which represent the conclusion, and include a sentence explaining how the truth table supports your answer. Your explanation should show that you understand what is means for a form of argument to be valid or invalid. Argument form: P→ r q→ r ..pvqrDetermine the truth value of each compound statement when p is false, q is false, and r is true. You must show how you arrived at your answer for each question. (∼p→r)↔(p∨∼q)(∼p→r)↔(p∨∼q) ∼[p→(q∧r)]∼[p→(q∧r)] ∼[(∼p→r)↔(p∨∼q)]∼[(∼p→r)↔(p∨∼q)] (∼p↔r)∧(∼q↔r)(∼p↔r)∧(∼q↔r) (q∧∼p)↔∼r1. With the aid of Truth tables, determine which of the following logical statements (or expressions)is a Tautology, a Contradiction, or a Contingency.(a) (p ↔ q) ↔ ((p → q) ∧ (q → p))(b) (p → (q → r)) ↔ (p ∧ q → r)(c) (p → r ∨ q)⊕(q ∧ r)(d) p ∨ q ∧ (¬r → ¬p ∧ ¬q) 2. With reference to the following logical equivalences, determine whether each of them is valid orinvalid. If it is invalid then give a counterexample (e.g. based on a Truth-Value assignment).If it is valid then give an algebraic proof using ONLY logical equivalences from Table 6 inSection 1.3 of course textbook, and the rule for conditional: p → q ≡ ¬p ∨ q.(a) p → (q → r) ≡ (p → q) → r(b) (p → r) ∨ (q → r) ≡ (p ∧ q) → r(c) (¬p → (q ∧ ¬q)) ≡ p(d) p ∧ q → r ≡ (¬p ∧ ¬q) → ¬r 3. Express the following logical statements in terms of the Contrapositive, Converse, and Inverse,respectively.(a) You are an adult provided that you are above 18 years of age.(b) It is daylight whenever the sun is shining.(c) 3 × 2 = 6 only if 2 × 3 = 6.
- Construct the truth table for the compound proposition ¬(p ∨ q) ↔️ q Use as many rows and columns as necessary. When you create your table, don't forget the header to label the columns. Include all subexpressions ( ex: before you can solve ¬(p ∨ q), you need a column for p ∨ q ).Consider the following case. What is the answer for question c, d, and e?5. Check whether p →r is a valid conclusion from the following premises: p → q V ¬r, q → p^r 6. Using indirect method, prove that q is a valid conclusion from the following premises: p → q,r → q, s → (p V r) 7. Using conditional proof, show that ¬q → s can be derived from the following premises: p → (q → r),¬r V p, q
- 1. How many rows appear in a truth table for each of these compound propositions? a) (q →¬p) v (¬p → ¬q) b) (p v ¬t) ^ (p v ¬s) c) (p→r) v (¬s → ¬t) v (¬u → v) d) (p ar as) v (q nt) v (r ^¬ t)Express the following in canonical forms , simplify using KMAP and make a truth table of the simplified expression: 2. F(A, B, C, D) = ∑ (0, 1,2,5,8,9, 10) Note: Pls. show groupings and highlight if possible.Consider the following NDFSM M, list the first ten elements of L(M) in lexicographic order (shortest first, then alphabetically if same length). b E, a, aa, ab, aaa, aab, baa, bab, aaaa, aaab a, aa, ab, aaa, aab, baa, bab, aaaa, aaab, aaba E, a, aa, ba, aaa, aba, baa, bba, aaaa, aaba a, aa, ba, aaa, aba, baa, bba, aaaa, aaba, aabb