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- Determine whether the following proposition is a tautology: (¬p∨¬(r⟶q))⟷(p⟶(¬q∧r)) can i get a non handwriting answer so it would be easy to copy please) Show that ∀xP(x) ∧ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∧ P(y)) The quantifiers have the same non empty domain. I know that to prove a proposition is logically equivalent to another one, I have to show that ∀xP(x) ∧ ∃xQ(x) ↔ ∀x∃y(P(x) ∧ P(y)) Which means I have to prove that (∀xP(x) ∧ ∃xQ(x)) → ∀x∃y(P(x) ∧ P(y)) ∧ ∀x∃y(P(x) ∧ P(y)) → (∀xP(x) ∧ ∃xQ(x)) I don't know the answer, so I saw the textbook answer. It says (1) Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. I get this part. Because P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ P(y)) is true. Emm... I think ∀x∃y(P(x) ∧ P(y)) is true because ∀x only affects P(x) and ∃y only affects P(y) since their alphabets are different. So, it has the exact same meaning as ∀xP(x) ∧ ∃yQ(y). And since the domains are the same, ∀xP(x) ∧ ∃yQ(y) is actually equal to ∀xP(x) ∧ ∃xQ(x). But the textbook states that "P(x) ∧ Q(x) is…Subject : calculas Show that: ¬q 1) p→¬q 2) (p∧r)∨s 3) s→(t∨u) 4) ¬t∧¬u where ¬ is denied.
- Logical equivalence of the given proposition: p if and only if q and (p→g) ^ (q→p)5. Let R = {1, 3, π, 4, 1, 9, 10}, S = {{1}, 3, 9, 10}, T = {1, 3, π}, and U = {{1, 3, π}, 1}. Which of thefollowing are true? For those that are not, why not? (d) 1 ⊆ U(e) {1} ⊆ T(f) {1} ⊆ S(g) {1} ∈ SProve that (p →(q →r)) → ((p →q) →(p →r)) ≡ T for all propositions p, q, r.
- Prove or disprove that the two propositions in each pair are equivalent. (p (q r)) , ((p q ) ( p r )) (p (q r)) , (( p q ) ( p r )) (( p q ) r ) , ( p ( q r))Answer the following: This problem exercises the basic concepts of game playing, using tic-tac-toe (noughts and crosses) as an example. We define Xn as the number of rows, columns, or diagonals with exactly n X’s and no O’s. Similarly, On is the number of rows, columns, or diagonals with just n O’s. The utility function assigns +1 to any position with X3=1 and −1 to any position with O3=1. All other terminal positions have utility 0. For nonterminal positions, we use a linear evaluation function defined as Eval(s)=3X2(s)+X1(s)−(3O2(s)+O1(s)). a. Show the whole game tree starting from an empty board down to depth 2 (i.e., one X and one O on the board), taking symmetry into account. b. Mark on your tree the evaluations of all the positions at depth 2. c .Using the minimax algorithm, mark on your tree the backed-up values for the positions at depths 1 and 0, and use those values to choose the best starting move. Provide original solutions including original diagram for part a!Answer the following: This problem exercises the basic concepts of game playing, using tic-tac-toe (noughts and crosses) as an example. We define Xn as the number of rows, columns, or diagonals with exactly n X’s and no O’s. Similarly, On is the number of rows, columns, or diagonals with just n O’s. The utility function assigns +1 to any position with X3=1 and −1 to any position with O3=1. All other terminal positions have utility 0. For nonterminal positions, we use a linear evaluation function defined as Eval(s)=3X2(s)+X1(s)−(3O2(s)+O1(s)). a. Show the whole game tree starting from an empty board down to depth 2 (i.e., one X and one O on the board), taking symmetry into account. b. Mark on your tree the evaluations of all the positions at depth 2. c .Using the minimax algorithm, mark on your tree the backed-up values for the positions at depths 1 and 0, and use those values to choose the best starting move. Provide original solution!
- Consider the wffs:φ1 ≡ p1 → (p2 → (p3 → p4))φ2 ≡ (p1 ∧ p2 ∧ p3) → p4(a) Technically speaking, neither φ1 nor φ2 is well-formed since neither is allowed by the formal syntaxof propositional logic. Correct them. Note, however, that we will freely make such trivial ’errors’ throughout this semester (as do most such courses).(b) Use truth tables (in the form defined in this course) to show that φ1 ↔ φ2.(c) After internalizing an intuitive understanding of this equality, propose an extension of it to n atoms.(d) State the number of rows in a truth table for proving the extensionConsider the wffs:φ1 ≡ p1 → (p2 → (p3 → p4))φ2 ≡ (p1 ∧ p2 ∧ p3) → p4(a) Technically speaking, neither φ1 nor φ2 is well-formed since neither is allowed by the formal syntaxof propositional logic. Correct them. Note, however, that we will freely make such trivial ’errors’throughout this semester (as do most such courses).(b) Use truth tables (in the form defined in this course) to show that φ1 ↔ φ2.(c) After internalizing an intuitive understanding of this equality, propose an extension of it to natoms.(d) State the number of rows in a truth table for proving the extension.Consider the following propositions: p="It is cloudy" q="It is rainy" r="I will play outside" s="I will stay at home" Select the inverse (s) of the follwing sentence: "I will stay at home and I will not play outside unless it is neither cloudy nor rainy"