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АВ' + AC + (ВC) + А A'B (A' + B) (B’ + B) C+ B'C' АВ'+ A(B' + С') + в (В' + С) AʼBC + AB'C' +A'B°C' +AB°C + ABC

АВ' + AC + (ВC) + А A'B (A' + B) (B’ + B) C+ B'C' АВ'+ A(B' + С') + в (В' + С) AʼBC + AB'C' +A'B°C' +AB°C + ABC

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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solve this and I provide some examples on guide you how to do it (pls pls follow the attachment i gave)

14. NOT OR (NOR) A+B Boolean Expression Equivalent Switching Circuit Boolean Algebra Law or Rule 15. Exclusive-OR A•B+A•B A+1=1 Annulment 16. Exclusive-NOR A•B+A• B A+0 = A Identity A.1 = A Identity Examples 1. A+ AB = A A.0 =0 Annulment Solution: A + AB А (1+ в) А (1) Factoring A out of both terms Applying Annulment law A+1 = 1 Applying Identity law 1A = A A+A= A Idempotent A.A=A Idempotent A |NOT A = A A+A=1 Double Negation Complement 2. A +AB= A+B Solution: Applying the previous rule to expand A term A + AB = A Factoring B out of 2nd and 3rd terms Applying Complement law A + A= 1 Applying identity law 1A = A A+ AB A.Ã = 0 Complement A+ AB + AB А-В (А+ A) A +B (1) A +B A+B=B+ A Commutative A•B=B. A Commutative 3. (A+ B) (A + C) = A + BC Solution: (A +B) (A + C) AA + AC + AB + BC Distributing tems Applying Idempotent law AA=A Factoring A out of 1* and 2nd terms Applying Annulment Law A+1 = 1 Applying Distributive Law Applying Annulment Law A+1 = 1 Applying Identity Law A. 1= A de Morgan's Theorem de Morgan's Theorem A+B =A•B A•B= A+B А+ АС + АВ + ВС А (1-C) + АВ + Вс A (1) + AB +BC А (1+B) + Вс A (1) +BC A+ BC Boolean Algebra Functions Function Description Expression 1. NULL 2. IDENTITY 1 Input A Input B 3. A 4. АВ + А (В-с) + В (В-C) Solution: 4. B 5. NOT A A АВ + A (В-C) + В (B-C) АВ + АВ + АC + ВB + ВС Apply Factoring Apply Idempotent Law BB = B Apply idempotent law A+ A =A Apply the rule in the 1ª example B+ BC =B Apply the rule in the 1" example B + AB = B NOT B A AND B (AND) 6. в 7. A•B AB + AB +AC +B+ BC AB + AC + B+ BC AB + AC + B B+ AC 8. A AND NOT B A•B 9. NOT A AND B A•B NOT AND (NAND) A ORB (OR) 10. A• B 11. A+B 12. A OR NOT B A+B 13. NOT A OR B В +В
Transcribed Image Text:14. NOT OR (NOR) A+B Boolean Expression Equivalent Switching Circuit Boolean Algebra Law or Rule 15. Exclusive-OR A•B+A•B A+1=1 Annulment 16. Exclusive-NOR A•B+A• B A+0 = A Identity A.1 = A Identity Examples 1. A+ AB = A A.0 =0 Annulment Solution: A + AB А (1+ в) А (1) Factoring A out of both terms Applying Annulment law A+1 = 1 Applying Identity law 1A = A A+A= A Idempotent A.A=A Idempotent A |NOT A = A A+A=1 Double Negation Complement 2. A +AB= A+B Solution: Applying the previous rule to expand A term A + AB = A Factoring B out of 2nd and 3rd terms Applying Complement law A + A= 1 Applying identity law 1A = A A+ AB A.Ã = 0 Complement A+ AB + AB А-В (А+ A) A +B (1) A +B A+B=B+ A Commutative A•B=B. A Commutative 3. (A+ B) (A + C) = A + BC Solution: (A +B) (A + C) AA + AC + AB + BC Distributing tems Applying Idempotent law AA=A Factoring A out of 1* and 2nd terms Applying Annulment Law A+1 = 1 Applying Distributive Law Applying Annulment Law A+1 = 1 Applying Identity Law A. 1= A de Morgan's Theorem de Morgan's Theorem A+B =A•B A•B= A+B А+ АС + АВ + ВС А (1-C) + АВ + Вс A (1) + AB +BC А (1+B) + Вс A (1) +BC A+ BC Boolean Algebra Functions Function Description Expression 1. NULL 2. IDENTITY 1 Input A Input B 3. A 4. АВ + А (В-с) + В (В-C) Solution: 4. B 5. NOT A A АВ + A (В-C) + В (B-C) АВ + АВ + АC + ВB + ВС Apply Factoring Apply Idempotent Law BB = B Apply idempotent law A+ A =A Apply the rule in the 1ª example B+ BC =B Apply the rule in the 1" example B + AB = B NOT B A AND B (AND) 6. в 7. A•B AB + AB +AC +B+ BC AB + AC + B+ BC AB + AC + B B+ AC 8. A AND NOT B A•B 9. NOT A AND B A•B NOT AND (NAND) A ORB (OR) 10. A• B 11. A+B 12. A OR NOT B A+B 13. NOT A OR B В +В
АВ' + AC + (ВСY + A А'B (A' + B) (В' + B) C+ B'C' АВ' + A(B' + C') + B (В' + C) A'BC + AB'C' +A'B'C' +AB'C + ABC
Transcribed Image Text:АВ' + AC + (ВСY + A А'B (A' + B) (В' + B) C+ B'C' АВ' + A(B' + C') + B (В' + C) A'BC + AB'C' +A'B'C' +AB'C + ABC
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