Intuition What this page is forThe parent note told you the three moving-furniture rules . This page does the drills . We will build a small table of every kind of situation these laws can throw at you — then knock down each cell with a fully worked example. When you finish, no exam variant should surprise you.
Reminder of the only two facts everything rests on:
A ⋅ B = min ( A , B ) A + B = max ( A , B ) 0 1
Think of every problem this topic can hand you as landing in one cell of the table below. Each row is a class of situation; the worked examples afterward are each tagged with the cell they cover.
#
Case class
What makes it tricky
Covered by
C1
Pure commutative re-order
spotting a hidden common factor
Example 1
C2
Pure associative re-group
2-input gates building a 3-input function
Example 2
C3
Distributive forward (AND over OR expand)expanding to reveal cancellation
Example 3
C4
Distributive reverse (factoring)fewest-gate goal
Example 4
C5
OR-over-AND weird form (A + B C the duplicated term
Example 5
C6
Degenerate input (a variable forced to 0 1 law still holds, but collapses
Example 6
C7
All same (A + A A ⋅ A idempotent edge, easy to mis-simplify
Example 7
C8
Chained laws (comm → assoc → dist together)order of moves
Example 8
C9
Real-world word problem translate English → algebra
Example 9
C10
Exam twist (a trap that looks like a law but isn't)knowing when a law does NOT apply
Example 10
We verify every numeric/logical claim below in the checker. Let's go.
Rewrite C ⋅ A + B ⋅ A
Forecast: guess which variable is common, and which side each swap happens on, before reading.
Look at term 1: C ⋅ A A ⋅ C Why this step? Commutative for AND: C ⋅ A = A ⋅ C min ( C , A ) = min ( A , C ) A
Look at term 2: B ⋅ A → A ⋅ B Why this step? Same law, so both terms now start with A
Result: A ⋅ C + A ⋅ B Why this step? Now A
Verify: truth-table both against the original. For A = 1 , B = 0 , C = 1 = 1 ⋅ 0 (no) … C ⋅ A + B ⋅ A = 1 ⋅ 1 + 0 ⋅ 1 = 1 A ⋅ C + A ⋅ B = 1 ⋅ 1 + 1 ⋅ 0 = 1
You only own 2-input OR gates but need a 3-input OR of A , B , C
Forecast: will OR(OR(A,B),C) and OR(A,OR(B,C)) ever disagree? Guess yes/no.
Write build-1: ( A + B ) + C Why this step? This is "combine A , B C
Write build-2: A + ( B + C ) Why this step? "Combine B , C
Associative law says ( A + B ) + C = A + ( B + C ) = max ( A , B , C ) Why this step? max
Verify: hardest row is A = 0 , B = 1 , C = 0 ( 0 + 1 ) + 0 = 1 + 0 = 1 0 + ( 1 + 0 ) = 0 + 1 = 1 1 ( 0 + 0 ) + 0 = 0 = 0 + ( 0 + 0 )
Simplify A ⋅ ( A + B )
Forecast: guess the final single-letter answer before working.
Expand with AND-over-OR: A ⋅ ( A + B ) = A ⋅ A + A ⋅ B Why this step? Distributive spreads the outer A
Use A ⋅ A = A Boolean Algebra Identity & Null Laws ): = A + A ⋅ B Why this step? min ( A , A ) = A
Factor A A + A ⋅ B = A ⋅ ( 1 + B ) 1 + B = 1 = A ⋅ 1 = A Why this step? 1 + anything = 1 Absorption result.
Verify: row A = 1 , B = 0 = 1 ⋅ ( 1 + 0 ) = 1 ⋅ 1 = 1 A = 1 A = 0 , B = 1 = 0 ⋅ ( 0 + 1 ) = 0 ⋅ 1 = 0 0
Simplify A ⋅ C + A ⋅ D + A ⋅ E
Forecast: how many gates does the raw form need? Guess, then count.
Spot the common A Why this step? Reverse distributive (factoring) only works when a variable appears in every product term.
Factor it out: A ⋅ C + A ⋅ D + A ⋅ E = A ⋅ ( C + D + E ) Why this step? Distributive run backwards pulls the shared A
Gate count. Before: 3 AND gates + 1 three-input OR = 4 After: 1 three-input OR + 1 AND = 2 Why this step? Fewer gates = cheaper, faster chip — the whole payoff (see Combinational Circuit Simplification ).
Verify: row A = 1 , C = 0 , D = 1 , E = 0 = 0 + 1 + 0 = 1 = 1 ⋅ ( 0 + 1 + 0 ) = 1 ⋅ 1 = 1 A = 0 C , D , E = 0 + 0 + 0 = 0 = 0 ⋅ ( … ) = 0 = 4 − 2 = 2
Expand X + ( Y ⋅ Z )
Forecast: how many times must X
Apply OR-over-AND distributive: X + Y ⋅ Z = ( X + Y ) ⋅ ( X + Z ) Why this step? This is the weird Boolean-only form. The lone X duplicates into both brackets — the single most-missed detail.
Sanity: does X X = 1 1 ( 1 + Y ) = ( 1 + Z ) = 1 = 1 Why this step? Confirms the duplication is not a typo — it is forced by the logic.
Verify: the trap row X = 1 , Y = 0 , Z = 0 = 1 + ( 0 ⋅ 0 ) = 1 + 0 = 1 = ( 1 + 0 ) ( 1 + 0 ) = 1 ⋅ 1 = 1 wrong single-copy form ( X + Y ) ⋅ Z = ( 1 + 0 ) ⋅ 0 = 0 X
In the expression A + ( B ⋅ C ) = ( A + B ) ( A + C ) C = 0
Forecast: guess what the whole thing collapses to once C = 0
Substitute C = 0 A + ( B ⋅ 0 ) = A + 0 = A Why this step? B ⋅ 0 = min ( B , 0 ) = 0 A + 0 = A
Substitute C = 0 ( A + B ) ⋅ ( A + 0 ) = ( A + B ) ⋅ A Why this step? A + 0 = A
Simplify ( A + B ) ⋅ A ⇒ A Why this step? Absorption — ANDing A ( A + B ) A
Both sides = A
Verify: A = 0 , B = 1 , C = 0 0 + ( 1 ⋅ 0 ) = 0 ( 0 + 1 ) ( 0 + 0 ) = 1 ⋅ 0 = 0 A = 1 , B = 1 , C = 0 1 + 0 = 1 ( 1 + 1 ) ( 1 + 0 ) = 1 ⋅ 1 = 1
A student writes A + A = 2 A A ⋅ A = A 2 A ⋅ A + A
Forecast: does 2 A A 2 exist in Boolean algebra? Guess before reading.
There is no 2 — values are only 0 1 A + A = max ( A , A ) = A A ⋅ A = min ( A , A ) = A Why this step? max min
Now simplify A ⋅ A + A = A + A = A Why this step? Both idempotent collapses fire, leaving a single A
Verify: A = 1 1 ⋅ 1 + 1 = 1 + 1 = 1 = A A = 0 0 ⋅ 0 + 0 = 0 = A 2 A 2 A 2 1
Simplify B ⋅ A + C + A ⋅ C
Forecast: guess whether the final form contains B
Commute term 1: B ⋅ A = A ⋅ B → A ⋅ B + C + A ⋅ C Why this step? Line up the A
Re-order (commutative on the OR chain) to group A A ⋅ B + A ⋅ C + C Why this step? OR is commutative, so we may sit the two A
Factor A A ⋅ ( B + C ) + C Why this step? Shared A
Expand back the other way — actually use OR-over-AND / absorption: note A ( B + C ) + C C A ( B + C ) + C = A B + A C + C A C + C = C ( A + 1 ) = C Why this step? A + 1 = 1 A C + C C
So expression = A B + C Why this step? Only the A B C B does remain (inside A B
Final: A ⋅ B + C
Verify: row A = 1 , B = 1 , C = 0 1 ⋅ 1 + 0 + 1 ⋅ 0 = 1 + 0 + 0 = 1 1 ⋅ 1 + 0 = 1 A = 0 , B = 1 , C = 1 0 + 1 + 0 = 1 0 + 1 = 1 A = 1 , B = 0 , C = 0 0 + 0 + 0 = 0 0 + 0 = 0
A door unlocks if: (the master key is inserted) OR (a valid card is present AND the PIN is correct) . Let M V P
Forecast: guess how many brackets the product-of-sums form has.
Translate: unlock U = M + ( V ⋅ P ) Why this step? "OR" = + ⋅
Apply OR-over-AND distributive: M + V ⋅ P = ( M + V ) ⋅ ( M + P ) Why this step? Some gate libraries are cheaper with OR-then-AND; this is the equivalent build. M
Read it back in English: unlock if (master OR valid card) AND (master OR PIN correct) .
Why this step? Confirms the meaning is preserved — master alone still opens everything.
Verify: master alone M = 1 , V = 0 , P = 0 = 1 + 0 = 1 = ( 1 + 0 ) ( 1 + 0 ) = 1 M = 0 , V = 1 , P = 0 = 0 + ( 1 ⋅ 0 ) = 0 = ( 0 + 1 ) ( 0 + 0 ) = 1 ⋅ 0 = 0
True or false: ( A + B ) ⋅ C = A + ( B ⋅ C ) correct relationship.
Forecast: guess true/false before touching a truth table.
False. Associativity only re-groups one operator at a time; here we have + and ⋅ Why this step? Crossing operators is the distributive law's job, not associativity's.Counterexample A = 1 , B = 0 , C = 0 ( 1 + 0 ) ⋅ 0 = 1 ⋅ 0 = 0 1 + ( 0 ⋅ 0 ) = 1 + 0 = 1 0 = 1 not equal .
Why this step? One mismatched row is enough to kill a claimed identity.
The correct expansion of ( A + B ) ⋅ C ( A + B ) ⋅ C = A ⋅ C + B ⋅ C Why this step? AND-over-OR spreads C is a valid law.
Verify: at A = 1 , B = 0 , C = 0 A ⋅ C + B ⋅ C = 0 + 0 = 0 0 A = 0 , B = 1 , C = 1 ( 0 + 1 ) ⋅ 1 = 1 0 ⋅ 1 + 1 ⋅ 1 = 1 A + B C = 0 + 1 = 1
Recall Cover the answers and speak them aloud
Factor A ⋅ P + A ⋅ Q + A ⋅ R
Expand M + V ⋅ P
Why is ( A + B ) ⋅ C = A + ( B ⋅ C )
Simplify A ⋅ ( A + B )
Answer to 1 A ( P + Q + R )
Answer to 2 ( M + V ) ( M + P ) M
Answer to 3 False; it mixes + ⋅ ( A + B ) C = A C + B C
Answer to 4 A A ( A + B ) = A ⋅ A + A ⋅ B = A + A B = A ( 1 + B ) = A
Mnemonic The one-line survival kit"Factor when a letter is in every term; duplicate the lone letter across both OR-brackets; never associate across a mixed operator — that's distributive's job."
Parent: the three laws Boolean Algebra Identity & Null Laws (the A ⋅ A = A A + 0 = A 1 + X = 1 De Morgan's Theorems (the next simplification tool once these run out)Logic Gates AND OR NOT (what the ⋅ + Karnaugh Maps (automates the factoring you did by hand here)Combinational Circuit Simplification (where the gate-savings pay off) 