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I want to simplify this logic expression:

Y = (A ∧ B ∧ ¬C ∧ D ) ∨ (C ∧ ¬D) ∨ (A ∧ B ∧ C) ∨ (¬A ∧ C)

I know it must become Y = (A ∧ B ∧ D) ∨ (C ∧ ¬D) ∨ (¬A ∧ C) and I found it with Karnaugh, but I can't find it with boolean simplification. I arrive here:

Y = (A ∧ B ∧ C) ∨ (A ∧ B ∧ D) ∨ (¬A ∧ C) ∨ (C ∧ ¬D)

Can anyone help me with this, explaining me how to arrive to the solution? Thanks!

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  • $\begingroup$ Wolfram Alpha can help you here, at least tell you what your goal should be, and it is not the $\;A \land B \land C\;$ that the first answer (correctly) deduced from your incorrect intermediate result. $\endgroup$
    – MarnixKlooster ReinstateMonica
    Dec 11, 2014 at 5:55
  • $\begingroup$ I know Wolfram, infact I found from it the correct final result I've written, but I still don't know how to find it. $\endgroup$
    – Vitto
    Dec 11, 2014 at 17:05

1 Answer 1

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$(A \land B \land C) \lor (A \land B \land D) $

can be simplified to

$A \land B \land (C \lor D) $

which can be rewritten as:

$A\land B\land ((C\land D)\lor(C\land \lnot D) \lor (\lnot C \land D))$

Notice that $C \land \lnot D$ appears later in Y, so we ignore that term (since Y would then be true anyway):

$A\land B \land ((C\land D) \lor (\lnot C\land D))$

which finally simplifies to

$A \land B \land D$

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  • $\begingroup$ Thank you, but I am not convinced about ignoring C∧¬D, because it appears later, but here it's inside brackets $\endgroup$
    – Vitto
    Dec 11, 2014 at 17:08
  • $\begingroup$ If $C\land\lnot D$ is true then Y is true, and so we wouldn't care about $A \land B$. $\endgroup$
    – sbares
    Dec 11, 2014 at 18:25
  • $\begingroup$ I rewrote it and I understood, thank you so much! $\endgroup$
    – Vitto
    Dec 11, 2014 at 18:41

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