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I have this circuit where I want to draw it's truth table. In fact I am new to this, I used to have the truth table and make the circuit. Can someone give me hand in how I should do it?
I drew the truth table with the inputs and since I have 4 inputs and a 4-to-1 MUX I should divide the output of the truth table in 4 groups. If I had only the MUX it would have been easy for me, but putting the decoder there, things got a bit complicated there.

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This is my truth table and that is what I meant by dividing the output(neglect the 0's). Each part of the output, I guess, is dependent on each input of the decoder. So any help would be appreciated. Thank you.

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    \$\begingroup\$ Do you know how the decoder works? Just add four columns for O0…O3. \$\endgroup\$ – CL. May 28 at 11:41
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So I figured out the answer:
O0 = A'B'
O1 = A'B
O2 = AB'
O3 = AB

When the selection input of the multiplexer is 00, we have 0 => O0 so we do A'B' whenever we have 00 in the truth table for CD and so on...

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  • \$\begingroup\$ Tip: use 2 spaces at the end of a line to force a line break. \$\endgroup\$ – Transistor May 28 at 12:36
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    \$\begingroup\$ This answer is incomplete because it doesn't follow up with the logic of the multiplexer and what combinations give F. If you are going to answer your own question you need to think about what other folk will think when they come to read it in a month or a year's time. An answer that seems plenty for you doesn't necessarily make it a good answer. \$\endgroup\$ – Andy aka May 28 at 13:15

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