2oo4 Logic (2 out of 4)

So as to not confuse the OP, the outputs of all the AND gates shown need to feed into one big OR gate (or a network of cascaded OR gates functioning as a 6-input OR gate). Here is an example using all 2-input logic gates:

Oops -- I was a little too hasty with my diagram. Here's a simpler version with one less OR gate required:

The table shows the required combinations: arrangements of 4 taken 2 =4*3/(1*2)=6The scheme copies the table. OR-gates with 3 entries are available as 4 units/chip.It can be build with 2 chips. I was not careful enough previous time I hope to be this one.

The condition can be also written as below (if I am not wrong):

"If a or b and c or d are high then output is high".

You need 2 "or - gates" and 1 "and - gate" = 3 gates.

The less components you use the less problem you can have.

What about [a and b] or [c and d]?

Unfortunately, this logic will not cover the cases [a and c] or [b and c] or [a and d] or [b and d]. A true 2oo4 system must work on *any* two inputs out of four.