What is the result of the following Boolean expression? A xor A xor A The truth table for exclusive OR operation: 2 0 0 1 1 0 1 1 A O A or (not B) not B not A B A xor B 0 1 1 O
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- Build truth tables or abbreviated truth tables (on your own scratch paper) to determine whether the following arguments are valid or invalid. (A v B) --> (C • D), (X v ~Y) --> (~C • ~W), (X v Z) --> (A • E), ∴∴~X Group of answer choices A. Valid B. InvalidComplete the truth table for the implication. You must submit a complete TT. (A ∧ ~B) → C. A B C (A ∧ ~B) → C T T T T T F T F T T F F F T T F T F F F T F F FPlease help me make a Truth Table on this F = A'B'C' + A'BC + AB'C + ABC' + ABC please make like this in my example attached photo thank youu!
- What is the outcome of the following given that isAwake = true, num1 = 7, num2 = 56, name = "John" and specialSymbol = "o"; isAwake || num1 < 5 Select one: True FalseUse a truth table to demonstrate that these two statements are logically equivalent:((p ∧ ¬q) ∨ (p ∧ q)) ∧ q and (p ∧ q).Please help me make a Truth Table on this F = A'B'C' + A'BC + AB'C + ABC' + ABC please make like this in my example attached photo.
- how this boolean expression works in what terms of the laws?: x'y'z' + x'y'z + x'yz + xy'z = x'y'(z+z') + z(x'y + xy') = x'y' + x'z + y'z I understood z+z' becomes 1 by the the inverse law, but how z(x'y + xy') goes x'z + y'z?not handwritten Consider CNF below. (D(y) ∨ E(y)) ∧ (D(y) ∨ ¬E(Miles)) ∧ ¬D(Carl) Variable y is universally quantified. Show that the expression always evaluates to false.PLEASE EXPLAIN ALL STEPSCreate a truth table of the equation:((((Y AND (NOT Z)) OR ((NOT Y) AND Z)) OR (NOT Z)) AND (NOT(X OR Y))) OR (NOT(((Y AND (NOT Z)) OR ((NOT Y) AND Z)) OR (NOT Z))) AND (X OR Y))
- Elimination of an arbitrary constant by erasure is a valid mathematical operation. True or FalseQuestion 4the complement of the following logical expression xy'+ywUsing the laws of boolean algebra simplify these expressions and LIST the laws used to justify each step. . Each final result should be expressed as a sum of products of literals, where each product uses the smallest number of literals. a. ABC + ABC’ + A’C + A’B’C + AB’C b. A'B' + A'BC' + (A + C')' c. A' + A'B'CD' + A'B'C'D' + AB'C' + AB'CD' + ABD + BC'D