Boolean Algebra, removal of redundant term 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.
 A: 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.

Distributive law
De Morgan's law
(I also used the associative law implicitly throughout.)

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

A: $A\neg B$ +$BC\neg D$ 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
