1-bit full adder

1. To begin, consider how you do binary addition on paper, but only look at what happens in a single bit. Ultimately, for each bit, you have 3 inputs, and 2 outputs. The inputs are A and B, the two bits you are adding, along with a potential Carry In (Cin, some sort of carry from the previous bit). The outputs are the sum for that bit, and a Carry Out (Cout) that goes onto the next bit. Observe the diagram above for clarification (Red = A, Light Blue = B, Green = Cin, Pink = Sum, Purple = Cout):
2. 
3. Consider all of the possible permutations of A, B, and Cin being 0 and 1, and fill out what Sum and Cout would be in the table below

   1. A	B	Cin	Sum	Cout
      0	0	0
      0	0	1
      0	1	0
      0	1	1
      1	0	0
      1	0	1
      1	1	0
      1	1	1

4. Now that you have a table for the 1-bit Full Adder, you should be able to directly come up with the sum of product (SOP) equation for Sum and Cout (separately, in terms of A, B, and Cin).
5. Use a K-map to minimize Cout (Sum would be requested to, but doesnt simplify, so just Cout).
6. Create 3 inputs, A, B, and Cin, and create 2 outputs, Sum and Cout
7. Implement a Full adder using any method you would like, but you cannot use the Adder component under components ----> arithmetic.
   1. You can either use the SOP from the truth table to make the schematic, or use the XOR
   2. Note: If you make the schematic from SOP, because this table has 2 outputs, you essentially need to make two circuits in one. For doing Sum, ignore Cout and implement Sum in terms of A, B, and Cin. For doing Cout, ignore Sum and implement Cout in terms of A, B, and Cin (but remember to use a K-map to simplify Cout and make the schematic from the simplified form).

1 0 0 1 1 + 0 1 0 1 1 000 C 1 0 0 1 1 + 0 1 0 1 1 000 C