Judging from the Karnaugh map, you can't get fewer than 5 conjunctions of literals. It seems that the only way to make the formula shorter is to consider not only DNFs.
I want to simplify this boolean expression
and I getCode:(!D^!C^!B^A)v(!D^!C^B^!A)v(!D^!C^B^A)v(!D^C^!B^A)v(!D^C^B^A)v(D^!C^B^A)v(D^C^!B^A)
I am pretty sure it can be further simplified, does anyone see a way?Code:( A^!D^!B ) v ( A^!D^C ) v (!B^A^C) v (!C^B^!D) v (!C^B^A)
And the order of the variables has to be D, C, B, A, where A is the least significant bit.