MHB A function given by a logical expression write in the truth table.

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The logical expression Y = A.B.D + A.not(C).D + A.not(B).C.D + A.D is evaluated in the truth table for all combinations of inputs A, B, C, and D. The output Y is 1 when A is 1 and D is 1, regardless of B and C's values. The truth table reflects that Y is 0 for all other combinations of inputs. This indicates that the function primarily depends on the values of A and D. Thus, the simplified expression for Y can be represented as Y = A.D.
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Assignment:
A function given by a logical expression Y = A.B.D + A.not (C) .D + A.not (B) .C.D + A.D write in the truth table.
 
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A B C D Y0 0 0 0 00 0 0 1 00 0 1 0 00 0 1 1 00 1 0 0 00 1 0 1 00 1 1 0 00 1 1 1 01 0 0 0 01 0 0 1 01 0 1 0 01 0 1 1 11 1 0 0 01 1 0 1 11 1 1 0 01 1 1 1 1
 


| A | B | C | D | Y |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |

In this truth table, A, B, C, and D represent different logical variables, and Y represents the output of the function. The function is evaluated for every possible combination of inputs, and the corresponding output is shown in the last column. Based on the given logical expression, the output Y will be 1 only when A is 1 and D is 1, regardless of the values of B and C. Therefore, the truth table shows that the function Y = A.D.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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