Tricky Problem: Simplifying F = AB + AB' + A'C

Apply the complement law and the absorption (elimination) law in sequence to fully minimize a three-term Boolean expression, instead of stopping after only the first simplification step.

Simplify the Boolean expression F = AB + AB' + A'C. This looks like a routine Boolean-algebra exercise, but it is designed to catch anyone who applies only one simplification law and declares victory too early — a full minimization here requires two separate laws applied back to back.

Why This Problem Is Tricky

F has three product terms: AB, AB', and A'C. The first two terms share the factor A and differ only in B versus B', which is the classic setup for the complement law (X + X' = 1). Many students correctly spot that and simplify AB + AB' down to A. The trap is stopping there: the resulting expression A + A'C still contains a second, less obvious reduction — A'C can be absorbed into A using a second law entirely, one that is easy to forget exists once the first simplification feels 'done.'

Recognizing that a Boolean simplification problem may require chaining two different laws, rather than a single pattern match, is exactly the skill this problem is built to test.

Step-by-Step Worked Solution

Start with F = AB + AB' + A'C. Group the first two terms, which share the common factor A: AB + AB' = A(B + B'). By the complement law, B + B' = 1, so A(B + B') = A · 1 = A. The expression is now F = A + A'C.

The remaining expression A + A'C matches the pattern X + X'Y, where X = A and Y = C. This pattern simplifies to X + Y by the absorption law (elimination form): X + X'Y = X + Y. Proof of this identity from first principles: X + X'Y = (X + X')(X + Y) [distributing OR over the AND-like structure, i.e. applying the distributive law in reverse] = 1 · (X + Y) = X + Y. Applying it here with X = A and Y = C gives A + A'C = A + C.

Therefore the fully minimized expression is F = A + C — a two-literal expression built from the original three three-literal-or-fewer terms, with the variable B eliminated entirely.

Why the Obvious (Incomplete) Answer Is Wrong

The most common wrong answer is F = A + A'C, submitted as if the problem were finished after only the complement-law step. It is not incorrect as a Boolean expression — it is logically equivalent to A + C — but it is not minimized, because it still contains a redundant literal (A') that a second, standard law removes. A problem that asks you to 'simplify' or 'minimize' is asking for the fewest literals and terms possible, not merely 'a' simpler form.

This matters in digital design because A + A'C would translate into an AND-OR-INVERT structure needing an extra AND gate and an extra inverter for A', while A + C needs only a single 2-input OR gate. Missing the second law costs real gates, real silicon area, and an extra layer of propagation delay.

The Two Laws Used, in Order

The complement law removes a variable that appears ANDed with its own complement across two terms sharing a common factor. The absorption/elimination form removes a variable that appears complemented in one term but is already 'covered' by a bare literal elsewhere in the expression. Both laws can be proven from the same three basic axioms (identity, distributive, complement), so neither is a shortcut — they are both fully rigorous.

1) Complement law:   X + X' = 1
   Used as:          A·B + A·B' = A·(B + B') = A·1 = A

2) Absorption law (elimination form):   X + X'Y = X + Y
   Used as:          A + A'C = A + C

Code Examples

The derivation runs in two clearly separable stages. Stage one (lines 1-3) eliminates B using the complement law after factoring A out of the two terms that share it. Stage two (line 4) eliminates the remaining A' using the absorption law, because A + A'C already has A covering the case where A is true, so the A' condition on the second term is redundant. Expected result: F = A + C (a 2-literal, single-OR-gate result)

F = AB + AB' + A'C
  = A(B + B') + A'C        [factor A out of the first two terms]
  = A(1) + A'C             [complement law: B + B' = 1]
  = A + A'C                [identity law: A·1 = A]
  = A + C                  [absorption law: X + X'Y = X + Y, with X=A, Y=C]

Gate Count Before vs. After Minimization

Fully minimizing before building the circuit turns a 5-gate structure into a single 2-input OR gate.

BEFORE (F = AB + AB' + A'C):            AFTER (F = A + C):

  A ──┬───[AND]──┐                        A ──┐
      │    │     │                             ├─[OR]── F
  B ──┘    │     ├─[OR]── F              C ──┘
  A ──┬───[AND]──┤
 B'──┘    │      │
 A'──┬───[AND]───┘
  C ──┘

  3 AND gates + 1 inverter + 1 3-input OR      1 OR gate only

Why Full Minimization Matters

In real logic synthesis, every literal that survives simplification becomes a physical gate input, contributing to chip area, switching power, and propagation delay. A designer (or a synthesis tool) that stops at the first applicable law rather than fully minimizing an expression ships a circuit with unnecessary gates — multiplied across millions of instances of similar logic in a large chip, incomplete simplification translates directly into wasted silicon and wasted power.

A + A'C is a correct but non-minimal intermediate result. It still matches the X + X'Y pattern, which always simplifies further to X + Y via the absorption law. Stopping after only the complement-law step leaves a redundant A' literal in the final answer.

Simplify F = AB + AB' + A'C to its minimal form. A. F = A + A'C B. F = A + C C. F = AB + C D. F = A'B + C

B. F = A + C — AB + AB' factors to A(B+B') = A by the complement law, leaving F = A + A'C. That expression matches the pattern X + X'Y (X=A, Y=C), which the absorption law simplifies to X + Y = A + C. Stopping at A + A'C (option 1) is the classic incomplete-simplification trap; options 3 and 4 introduce literals that never appear in the correctly reduced expression.

Which Boolean identity justifies simplifying the expression X + X'Y directly to X + Y? A. De Morgan's law B. The absorption law (elimination form) C. The distributive law alone, with no complement law involved D. The idempotent law (X + X = X)

B. The absorption law (elimination form) — X + X'Y = X + Y is the elimination/absorption identity. It can itself be proven using the distributive and complement laws together (X + X'Y = (X+X')(X+Y) = 1·(X+Y) = X+Y), but as a named law on its own it is the absorption law, distinct from De Morgan's law (which relates AND/OR to their complements) and from idempotence (which just removes duplicate identical terms).

Simplify G = XY + XY' + X'Z using the same two-law technique used above.

XY + XY' = X(Y + Y') = X. So G = X + X'Z, which matches X + X'Y (with Y replaced by Z) and simplifies via absorption to G = X + Z.

First apply the complement law to the two terms that share the factor X, then look for an absorption-law pattern in what remains.