(¬a∨c)¬(ac)(b∨¬c)bc упростить

(¬a∨c)¬(ac)(b∨¬c)bc = (¬a + c) ¬(ac) (b + ¬c)bc =

(¬a + c) (¬a + ¬c) (bbc + bc¬c) = (¬a + ¬a¬c + ¬ac + c¬c)(bc + 0) =

(¬a + ¬a¬c + ¬ac + c¬c)(bc + 0) = (¬a(1 + ¬c + c) + 0) bc = ¬abc