Simplify Boolean Expression $A\cdot \overline{B\cdot C+C\cdot A}\cdot B$. Where did I go wrong with the simplification

Question

Question

Simplify Boolean Expression $A\cdot \overline{B\cdot C+C\cdot A}\cdot B$

My Approach

Start
$A\cdot \overline{B\cdot C+C\cdot A}\cdot B$

De Morgan's Law
$A\cdot(\overline{B}+\overline{C})\cdot(\overline{C}+\overline{A})\cdot B$

Distribution
$A \cdot (\overline{C}+\overline{A}) \cdot B \cdot \overline{B}+A \cdot (\overline{C}+\overline{A}) \cdot B \cdot \overline{C}$

Using $A \cdot \overline{A}=0$
$0+A \cdot (\overline{C}+\overline{A}) \cdot B \cdot \overline{C}$

Using $A+0=A$
$A \cdot (\overline{C}+\overline{A}) \cdot B \cdot \overline{C}$

Distribution
$A \cdot B \cdot \overline{C} \cdot \overline{C}+A \cdot B \cdot \overline{C} \cdot \overline{A}$

Using $A+A=A$
$A \cdot B \cdot \overline{C} \cdot \overline{A}$

Using $A \cdot \overline{A}=0$
$0$

Hence, Final Answer is $0$

While Sites are giving me this as the correct simplification $AB\overline{C}$

Answer 1

Here is your mistake:

Distribution: A•B•$\overline{C}$•$\overline{C}$+A•B•$\overline{C}$•$\overline{A}$

A+A=A : A•B•$\overline{C}$•$\overline{A}$

No. The two terms are different. In fact, the second term contains $A\cdot \bar{A}$, and thus works out to $0$, but the first term simplifies to just $A\cdot B \cdot \bar{C}$ ... as that website correctly finds.

Answer 2

You have line $6$

6) $A•B•\overline{C}•\overline{C}+A•B•\overline{C}•\overline{A}$

Then one line 7: you claim as $A+ A = A$ that $A•B•\overline{C}•\overline{C}+A•B•\overline{C}•\overline{A}= A•B•\overline{C}•\overline{A}$

But $A•B•\overline{C}•\overline{C}$ and $A•B•\overline{C}•\overline{A}$ are not the same.

Instead use $A\cdot A = A$ to get that $\overline{C}•\overline{C} = \overline{C}$ to get:

7) $A•B•\overline{C}+A•B•\overline{C}•\overline{A}$

Then distribute:

8) $A•B•\overline{C}(1 + \overline A)$

9) $1 + A = 1 : A•B•\overline{C}\cdot 1$

10) $1\cdot A = A: A\cdot B\cdot \overline{C}$.

...or....

8) $A\overline A = 0: A\cdot B\cdot \overline{C} + 0$

9) $A + 0 = A: A\cdot B\cdot \overline{C}$

Answer 3

Analogous with sets:

• $A\cap((B\cap C)\cup(C\cap A))^{\complement}\cap B$
• $A\cap((B\cap C)^{\complement}\cap(C\cap A)^{\complement})\cap B$
• $A\cap((B^{\complement}\cup C^{\complement})\cap(C^{\complement}\cup A^{\complement}))\cap B$
• $A\cap((B^{\complement}\cap C^{\complement})\cup(B^{\complement}\cap A^{\complement})\cup (C^{\complement}\cap C^{\complement})\cup(C^{\complement}\cap A^{\complement}))\cap B$
• $(A\cap B^{\complement}\cap C^{\complement}\cap B)\cup(A\cap B^{\complement}\cap A^{\complement}\cap B)\cup(A\cap C^{\complement}\cap C^{\complement}\cap B)\cup(A\cap C^{\complement}\cap A^{\complement}\cap B)$
• $\varnothing\cup\varnothing\cup(A\cap C^{\complement}\cap C^{\complement}\cap B)\cup\varnothing$
• $A\cap C^{\complement}\cap B$