# Circuit Gate Reduction. Reduce the following logic circuit using only two input gates.

#### wooshj

Joined Sep 25, 2017
1
You may only use two input gates, AND, OR and NOT gates are only gates accepted.

I'm new to this stuff so I know its probably really simple but anyways I have a truth table, a representation (sum-of-products and product-of-sum), and I ended up making the gate circuit for the given representation.

Heres the representation: ~(A*B*C) + (~A*B*C) + (A*~B*C) + (A*B*~C), the * is AND, the + is OR.

I'm suppose to reduce this using only two input gates instead of the 5 that i'm using now. The representation I have is the one I came up with based on the truth table, not a given one on the homework so it might be different then what something else.

Anyways I've tried reducing this representation and I've come to the conclusion theres no way I can? I need the ~(A*B*C) because 1 1 1 = 0 and 0 0 0 = 1 and if you try pulling out letters and reducing it you just end up with something dumb like B(~AC + A~C), A(~BC + ~CB) etc..

Any ideas?

#### WBahn

Joined Mar 31, 2012
29,130
TS has stated that they solved the problem.

@wooshj: Please don't delete content when you are done with it. Forums are archival in nature and the purpose is lost if people delete their content once they are finished with it.

#### RBR1317

Joined Nov 13, 2010
709
For archival consideration:

There is a single input combination equal to (A⋅B⋅C) and the first term, ~(A⋅B⋅C), makes the Boolean expression TRUE for every other input combination, thereby making all of the other terms redundant.