Give three possible ways to express the following Boolean function with eight or fewer literals: F = A’BC’D + AB’CD + A’B’C’ + ACD’
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Give three possible ways to express the following Boolean function with
eight or fewer literals:
F = A’BC’D + AB’CD + A’B’C’ + ACD’
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- Simplify or reduce the Boolean function below into three literals: F(P,Q,R) = (P + Q’R’)(PQ+R)+PQR’+P’RLet α = (a + b)^∗ab(a + b)^∗. Give a regular expression equivalent to ∼ α in the following cases.Simplify the expressions as much as possible (either using the rules from the lecture or othermeans).(a) Σ = {a,b}(b) Σ = {a,b,c}Simplify the following Boolean expressions to a minimum number of literals: (e) (a + b + c')(a' b' + c) (f) a'bc + abc' + abc + a'bc'
- Simplify the following Boolean expressions to a minimum number of literals:(a) ABC +A'B +ABC' (b) x'yz + xz(c) ( x +y)'(x'+y') (d) xy + x(wz + wz')(e) ( BC'+A'D) (AB'+CD') (f) (a' + c') (a + b' + c')Simplify the following Boolean expressions to a minimumnumber of literals: a. xyz + x ′ y + xyz ′ b. x ′ yz + xz c. ( x + y ) ′ ( x ′ + y ′ )Simplify The Following Expression To A Minimum Number Of Literals. ab'cd'e+e'h'+abcd'e+acd+ acf'gh' + acde'
- Simplify the following Boolean expressions to a minimum number of literals: a ′ bc + abc ′ + abc + a ′ bc ′Simplify the following Boolean expressions to a minimum number of literals:a. (a + b+ c’)(a’b’ + c)b. (a’ + c’) (a + b’ + c’)1. In transforming a formula into an equivalent CNF, you can use the absorption laws to eliminate conjunctions within disjuntions, i.e. expressions such as p ˅ (q ˄ r) and (p ˄ q) ˅ r. True False 2. Let p, q, and r be propositional variables. Which of the following expressions would NOT be formulas in conjunctive normal form? (p ˄ ¬q) ˅ (¬r ˄ q ˄ p) ¬q p ˅ q ˅ ¬p p ˄ ¬p 3. Consider the propositional logic formula (p ˄ q) ˅ (r ˅ ¬s) From the options below, which one is the equivalent CNF? To determine the correct answer, transform the formula above into CNF using the steps learnt in this module. (p ˅ r ˅ s) ˄ (q ˅ r ˅ ¬s) (¬p ∨ r ∨ s) ∧ (¬q ∨ ¬r ∨ ¬s) (p ˅ r ˅ ¬s) ˄ (q ˅ r ˅ s) (p ˅ r ˅ ¬s) ˄ (q ˅ r ˅ ¬s)…
- Simplify the following Boolean functions to a minimum number of literals1) F1(A, B, C) = A′B′C′ + AB′C′ + AB′C2) F2(x, y) = (x + y)′(x′ + y′)′3) F3(a, b, c, d) = abc′d + a′bd + abcd4) F4(A, B, C) = A′B′ + A′BC′ + (A + C′)′5) F5(a, b, c, d) = a′b(d′ + c′d) + b(a + a′cd)if p and q are logical variables, which of the following is a tautology (i.e., always correct irrespective of specific value of variables) Select one: a. p → (q ∧ p) b. p ∨ (q → q) c. (p ∨ q) → q d. p ∨ (p → q)Translate each of these nested quantifications into an English statement that expresses a mathematicalfact. The domain in each case consists of all real numbers.a) ∃x∀y(x + y = y) b) ∀x∀y (((x ≥ 0) ∧ (y < 0)) → (x − y > 0))c) ∃x∃y(((x ≤ 0) ∧ (y ≤ 0)) ∧ (x − y > 0))d) ∀x∀y((x ≠ 0) ∧ (y ≠ 0) ↔ (xy ≠ 0))