Determine the validity of the following arguments using truth tables. Which confluence of truth values shows its invalidity? 1. A ⇒B B ⇒A ----------- ∴A ⇒B 2. (E ∨R ) ∧(E ∨T) R ∧T ------------ ∴E 3.

Exercise 1.4.4: Proving whether two logical expressions are equivalent. About Determine whether the following pairs of expressions are logically equivalent. Prove your answer. If the pair is logically equivalent, then use a truth table to prove your answer. (a) ¬(p ∨ ¬q) and ¬p ∧ q (b) ¬(p ∨ ¬q) and ¬p ∧ ¬q (c) p ∧ (p → q) and p → q (d) p ∧ (p → q) and p ∧ q

The following argument of TFL is invalid. ((D → B) ∧ C) ∴ ((B → C) → D)) Use a one-line truth table to show that the argument is invalid. Explain why the truth table shows that the argument is invalid. (1-2 sentences)

Please answer with explanations.

Set up truth tables for the statementsP ⇒ (P ⇒ Q), (P ⇒ Q) ⇒ Q, P ∧ (∼ Q), (∼ P) ∨ (∼ Q)Are any of these statements equivalent?

Racket code only please Using the logical operators, and, or, implies, not, and iff, and atoms representing truth values, we can make a list that would represent proposition: For instance: '(A and (not A)), or '((B iff (A or C)) implies (not (C implies A))) would represent the formal statements (?∧¬?) and ((? ≡(?∨?)→¬(? →?))). We use the following convention when we represent propositions in this way: for every operator – unary or binary – there corresponds at least one pair of parentheses defining its scope (and perhaps more). Thus, the lists ‘(A and not A) and (B iff A or C) are not permitted but ‘((A and ((not A))) and ‘(B iff (((A or C)))) are acceptable. Define Evaluate-WFF, which will take as input a formula of prepositional logic in which there are only truth values and no propositional variables, such as the following: (#t or (not #t)). Evaluate-WFF should return #t if the formula under the particular truth assignment evaluates to #t and false otherwise. By way of example,…

Prove the following predicate logic statements valid or invalid: A) ∀x(A(x) → B(x)) ∧ ∀x(A(x) ∨ ¬C(x)) ∧ ∃x(¬B(x)) → ¬∃x(¬A(x) ∧ ¬B(x) ∧ ¬C(x)) B) ∀x(A(x) → B(x)) ∧ ∃x(A(x) ∨ B(x)) → ∃x(A(x) ∧ B(x))