It amuses me that student problems are invariably "urgent". By contrast, we who work in the commercial world of course have infinitely long timescales for our projects :-)
Not being an academic I have never heard of Shannon's expansion theorem, but Shannon was a pretty bright guy and I have no reason to doubt that he invented such a thing.
It's useful to remember that you can write a multiplexer as a Boolean expression:
Mux = in0.Sel' + in1.Sel
represents a 2:1 mux selected by Sel, with inputs in0 and in1. To help with this Mux description, let's define that as a function M(in0,in1,Sel).
So we can re-think your expression as multiplexers...
F = C' + (A' + A.B').C = C' + C.M(1, B', A) = M(1, M(1, B', A), C)
One more little observation: B' = 1.B' + 0.B = M(1, 0, B)
Hey, I got one over on your prof! I don't need that OR gate at all!
F = M(1, M(1, B', A), C) = M(1, M(1, M(1, 0, B), A), C)
|\ |\ 1-| | |\ 1-| | | |----F 1-| | | |------| | | |------| | |/ 0-| | |/ | |/ | C | A B
ASCII-schematic with thanks, as usual, to Andy Weber's wonderful AACircuit program
Shannon's theorem refers to the MUX expansion as you already explained, but goes deeper into unique minterm expansion claims. Now try it- but wait a week so the OP cannot claim extra credit on his assignment.:-)
You have by DeMorgan's Theorem that (A.B.C)'=A'+B'+C' so the cofactor wrt C is A'+ B', and the cofactor wrt A of A'+ B' is B'- so the Shannon expansion is C'."1" + C.( A'."1"+ A.B'), and this agrees with your expression. Now use Actel's "logic WHEEL" function to map it-) and you are 10% of the way there-)
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