B bogla New Member May 13, 2018 #1 Here is a problem I'm trying to solve. Here is the given solution in my book. I don't understand that marked area in the solution. Need help to understand it to move ahead. Can anybody please explain that in a simpler way? Thanks
Here is a problem I'm trying to solve. Here is the given solution in my book. I don't understand that marked area in the solution. Need help to understand it to move ahead. Can anybody please explain that in a simpler way? Thanks
MikeMl Well-Known Member Most Helpful Member May 13, 2018 #2 Let me try to restate the problem: 1. By definition, A*B=AB + !A!B, where !=not (the * operator is sorta like ex-or) 2. Let C=A*B 3. Verify that A=B*C 4. Substitute A*B for C in the above equation, A=B*A*B, but also put () around the last two terms, so it looks like this: A=B*(A*B) 5. Now use the definition of the connective operator (equ 1) to expand equ 4: A=!B!(A*B) + B(A*B) ps 50 years as a logic designer, and I never heard of the "connective"
Let me try to restate the problem: 1. By definition, A*B=AB + !A!B, where !=not (the * operator is sorta like ex-or) 2. Let C=A*B 3. Verify that A=B*C 4. Substitute A*B for C in the above equation, A=B*A*B, but also put () around the last two terms, so it looks like this: A=B*(A*B) 5. Now use the definition of the connective operator (equ 1) to expand equ 4: A=!B!(A*B) + B(A*B) ps 50 years as a logic designer, and I never heard of the "connective"
B bogla New Member May 13, 2018 #3 MikeMl said: 5. Now use the definition of the connective operator (equ 1) to expand equ 4: A=!B!(A*B) + B(A*B) Click to expand... awesome. got this part. But stuck in the next step here. Please see this step. Is it a law ? which law ? or how can we do that step there ?
MikeMl said: 5. Now use the definition of the connective operator (equ 1) to expand equ 4: A=!B!(A*B) + B(A*B) Click to expand... awesome. got this part. But stuck in the next step here. Please see this step. Is it a law ? which law ? or how can we do that step there ?
MikeMl Well-Known Member Most Helpful Member May 13, 2018 #4 That transition can be shown with a simple two-term, four-square Veich diagram, aka DeMorgan's Theorem Last edited: May 13, 2018
That transition can be shown with a simple two-term, four-square Veich diagram, aka DeMorgan's Theorem
R Ratchit Well-Known Member May 14, 2018 #5 MikeMl said: That transition can be shown with a simple two-term, four-square Veich diagram, aka DeMorgan's Theorem Click to expand... Don't knock yourself out trying to help this multiple poster. Different userid, but same individual. https://www.electronicspoint.com/threads/validity-of-equation.288416/ Ratch
MikeMl said: That transition can be shown with a simple two-term, four-square Veich diagram, aka DeMorgan's Theorem Click to expand... Don't knock yourself out trying to help this multiple poster. Different userid, but same individual. https://www.electronicspoint.com/threads/validity-of-equation.288416/ Ratch