Boolean Algebra, removal of redundant term

Question

How do I simplify this boolean expression

A¬B + A¬C + BC¬D + A¬D

to

A¬B + A¬C + BC¬D

with boolean algebra?

The A¬D is redundant, I can see why it is when I examine the truth table for this expression, but I can't see how to simplify it.

Answer 1

$A\neg B$ +$BC\neg D$ simplifies to $A\neg B$ +$BC\neg D$+$AC\neg D$ by shannon expansion formula and $AC\neg D$+$A\neg C$ simplifies to $A\neg D$ again by shannon expansion

Answer 2

Start with your second expression $$\begin{align} A\neg B + A\neg C + BC\neg D &= (\neg B + \neg C)A + BC\neg D\;\;\text{distributive law}\\ &= (\neg (BC))A + BC\neg D\;\;\;\;\;\;\text{De Morgan's law}\\ &=\neg XA + X\neg D\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;X = BC \text{ for simplicity}\\ &=\neg XA(D+\neg D) + X\neg D(A+\neg A)\\ &=\neg XAD \;+\; \neg XA\neg D \;+\; X\neg DA \;+\; X\neg D\neg A\;\;\;(\dagger) \end{align}$$ where the last but one line uses $Y+\neg Y = 1$, and the last line $(\dagger)$ uses the distributive law.

Now expand the additional term $A\neg D$ from your first expression: $$A\neg D = (\neg X + X)A\neg D = \neg X A\neg D + X\neg DA.\;\;$$ This expansion matches the second and third terms from $(\dagger)$, and so adding it to the second expression makes no difference.

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Distributive law

De Morgan's law

(I also used the associative law implicitly throughout.)

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The truth table is given below for reference

abcd xad result
0000 000      0
0001 001      0
0010 000      0
0011 001      0
0100 000      0
0101 001      0
0110 100      1
0111 101      0
1000 010      1
1001 011      1
1010 010      1
1011 011      1
1100 010      1
1101 011      1
1110 110      1
1111 111      0