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Simplifying a Boolean Expression: Y=ABC+A′BC+AB′CY = ABC + A'BC + AB'C

Simplifying a Boolean Expression: Y=ABC+A′BC+AB′CY = ABC + A'BC + AB'C

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Sep 14, 2026

We want to simplify the Boolean function:

Y=ABC+A′BC+AB′CY = ABC + A'BC + AB'C

Using Boolean algebra, we can factor common terms and apply the complement/absorption relationships to reduce the expression to a simpler equivalent form.3

Key keyword concepts we’ll use: complement, distributive factoring, and the identity that A+A′=1A + A' = 1 (and thus A(B)+A′(B)=BA(B) + A'(B) = B).2

Footnotes

  1. Boolean algebra laws - Wikipedia - Summarizes identities like distributive law, complement, and absorption. ↩

  2. Complement (logic) - Wikipedia - Explains A′A' and complement relationships. ↩ ↩2

  3. Boolean algebra - Wikipedia (Basic identities section) - Includes A+A′=1A + A' = 1 and related simplifications. ↩ ↩2

Boolean Algebra Simplification (Distributive/Factoring) - Crash Course

Stepwise simplification overview

The expression has three product terms:

  1. ABCABC
  2. A′BCA'BC
  3. AB′CAB'C

Notice the first two terms share BCBC:

ABC+A′BC=BC(A+A′)ABC + A'BC = BC(A + A')

Then use A+A′=1A + A' = 1, giving:

BC(A+A′)=BCBC(A + A') = BC

So:

Y=BC+AB′CY = BC + AB'C

Finally factor CC:

Y=C(B+AB′)Y = C(B + AB')

Then use absorption: B+AB′=B(1+A′)=BB + AB' = B(1 + A') = B (since 1+A′=11 + A' = 1), giving:

Y=BCY = BC

Simplify $Y = ABC + A'BC + AB'C$

  1. 1
    Step 1

    Start with Y=ABC+A′BC+AB′CY=ABC + A'BC + AB'C. Group the first two terms: ABC+A′BC=BC(A+A′)ABC + A'BC = BC(A + A').

  2. 2
    Step 2

    Apply A+A′=1A + A' = 1, so BC(A+A′)=BCBC(A + A') = BC.

  3. 3
    Step 3

    Now Y=BC+AB′CY = BC + AB'C.

  4. 4
    Step 4

    Write Y=C(B+AB′)Y = C(B + AB').

  5. 5
    Step 5

    Reduce B+AB′B + AB': B+AB′=B(1+A′)=BB + AB' = B(1 + A') = B; hence Y=C⋅B=BCY = C\cdot B = BC.

Full algebraic derivation (clean write-up)

[ \begin{aligned} Y &= ABC + A'BC + AB'C \ &= BC(A + A') + AB'C \quad &(\text{factor } BC)\ &= BC(1) + AB'C \quad &(A + A' = 1)\ &= BC + AB'C \ &= C(B + AB') \quad &(\text{factor } C)\ &= C\big(B(1 + A')\big) \quad &(\text{factor } B)\ &= C(B\cdot 1) \quad &(1 + A' = 1)\ &= BC \end{aligned} ] Therefore, the simplified result is:

Y=BC\boxed{Y = BC}

Common factoring trick

When two terms differ only by a complemented/non-complemented variable (like ABCABC and A′BCA'BC), factor the shared parts (here, BCBC) to expose A+A′=1A + A' = 1 quickly. This often collapses multiple terms at once.2

Footnotes

  1. Complement (logic) - Wikipedia - Explains A′A' and complement relationships. ↩

  2. Boolean algebra - Wikipedia (Basic identities section) - Includes A+A′=1A + A' = 1 and related simplifications. ↩

Avoid “dropping” terms too early

You can’t remove AB′CAB'C just because it looks different—only identities like A+A′=1A + A' = 1 or absorption/distributive transformations justify eliminating structure. Always show the algebra step-by-step.2

Footnotes

  1. Boolean algebra laws - Wikipedia - Summarizes identities like distributive law, complement, and absorption. ↩

  2. Complement (logic) - Wikipedia - Explains A′A' and complement relationships. ↩

Original vs Simplified (term counts)

Compare number of product terms before and after simplification.

Quick checks & intuition

Boolean Simplification Workflow

Factor common terms

1

Group terms sharing literals (e.g., BCBC in ABC+A′BCABC + A'BC)."

Use complement identities

2

Replace A+A′A + A' with 11 when it appears as a factor."

Absorb redundant structure

3

Reduce forms like B+AB′B + AB' to BB using distributive/absorption."

Return final simplified form

4

Conclude Y=BCY=BC."

Boolean Algebra Micro-Drills

1 / 4
Question · Term

What is the identity for complements?

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Answer · Definition

keyword

Knowledge Check

Question 1 of 3
Q1Single choice

Simplify ABC+A′BCABC + A'BC.

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