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Problem:

Given that $$ A \cdot B = 0$$ and $$A + B = 1$$ use algebraic manipulation to prove that: $$ (A + C) \cdot( \overline A + B ) \cdot (B + C) = B \cdot C$$

Answer:
First, I will rewrite the left hand side without the operator $$ \cdot $$. That is, I will write $$ AB $$ to mean $$A \cdot B$$.

\begin{align*} (A + C)( \overline A + B ) (B + C) &= ( 0 + \overline A C + AB + BC ) (B + C) \\ (A + C)( \overline A + B ) (B + C) &= ( 0 + \overline A C + 0 + BC ) (B + C) = ( \overline A C + BC )( B + C) \\ (A + C)( \overline A + B ) (B + C) &= B( \overline A C + BC ) + C( \overline A C + BC ) \\ (A + C)( \overline A + B ) (B + C) &= \overline A BC + BC + C( \overline A C + BC ) \\ (A + C)( \overline A + B ) (B + C) &= \overline A BC + BC + \overline A C + BC \\ (A + C)( \overline A + B ) (B + C) &= \overline A BC + BC + \overline A C \\ (A + C)( \overline A + B ) (B + C) &= \overline A BC + \overline A C + BC \\ (A + C)( \overline A + B ) (B + C) &= \overline A C + BC \\ \end{align*} I think I am on the right track, but I do not know how to finish solving the problem.
Thanks,
Bob

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  • \$\begingroup\$ How did you go from (0+!AC+AB+BC) to (0+!AC+0+BC) on the right side of step 1 to step 2? \$\endgroup\$ Oct 6, 2019 at 20:59
  • \$\begingroup\$ @MichaelKaras We are given that AB is 0. \$\endgroup\$
    – Bob
    Oct 6, 2019 at 21:11

1 Answer 1

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This is a super long detour for a simple result:

  1. A·B = 0 implies that at most one of A, B is true
  2. A+B = 1 implies that at least one of A, B is true
  3. This implies that exactly one of A, B is true.
  4. This implies that \$\overline A = B\$.

Then

\begin{align} (A + C) \cdot( \overline A + B ) \cdot (B + C) &=(\overline B +C)\cdot(B+B)\cdot (B+C)\\ &=B(\overline B +C)(B+C)\\ &=(B\overline B + BC)(B+C)\\ &=(0+BC)(B+C)\\ &= BCB + BCC\\ &=BC\\ &\overset!=BC \end{align}

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