discrete.

STEP-BY-STEP LESSON · §6.3

Disproofs and Algebraic Proofs

Choose counterexamples for false laws and justified transformations for true ones.

Before you begin

Set Theory: Definitions and the Element Method of Proof

The statement x∈A concerns one object. A⊆B says that every element of A belongs to B.

A⊆B ⇔ ∀x(x∈A ⇒ x∈B)

X=Y means X⊆Y and Y⊆X. One inclusion still allows extra elements in the second set.

X=Y ⇔ (X⊆Y ∧ Y⊆X)
Properties of Sets

Fix an arbitrary x in a common universe U. Write p for x∈A, q for x∈B and r for x∈C. Set operations become logical operations on these three statements.

∩ ↔ ∧; ∪ ↔ ∨; complement ↔ ¬

An element is outside A∪B exactly when it is outside A and outside B. Likewise it is outside A∩B exactly when it is outside at least one of A,B.

(A∪B)ᶜ=Aᶜ∩Bᶜ; (A∩B)ᶜ=Aᶜ∪Bᶜ
Direct Proof and Counterexample I: Introduction

Integer n is even iff there is an integer k with n=2k. Evenness gives this representation, and the representation gives evenness.

n even ↔ ∃k∈ℤ: n=2k

For “every odd integer has an odd square”, testing 3,5,7 suggests a pattern but leaves infinitely many cases. Write an arbitrary odd integer as 2k+1 with integer k; no special value is assumed.

n=2k+1; k∈ℤ

Symbols

A∖B
elements of A that are not in B
U
fixed universe containing all sets’ elements under discussion
Aᶜ
complement U∖A within the fixed universe U
Definitions and notation for this topic
Universe and complement laws
Aᶜ=U∖A; (A∪B)ᶜ=Aᶜ∩Bᶜ; (A∩B)ᶜ=Aᶜ∪Bᶜ

Fix a universe U and subsets A,B⊆U. Complement means outside the set but inside U; both De Morgan identities use this same universe.

An element is outside the union exactly when it is outside both sets. Related laws are A∪Aᶜ=U, A∩Aᶜ=∅, and (Aᶜ)ᶜ=A.

Separate glossary example

If U={1,2,3,4}, A={1,2}, B={2,3}, then (A∪B)ᶜ={4}=Aᶜ∩Bᶜ.

Open glossary card
Distributive and difference laws
A∩(B∪C)=(A∩B)∪(A∩C); A∖B=A∩Bᶜ

For A,B,C⊆U, set identities permit replacing a set expression by an equal one. The dual distributive law swaps ∩ and ∪ throughout.

Each equals sign needs a valid law. Alternatively, fix an arbitrary x∈U and show that membership in the left and right sides is equivalent.

Separate glossary example

(A∩B)∪(A∩Bᶜ)=A∩(B∪Bᶜ)=A∩U=A.

Open glossary card
Simplify

Rewrite an expression in a simpler equivalent form, respecting its domain.

Distribute A∩ over the union and use B∪Bᶜ=U. The rewritten expression has the same value under the stated universe; this is not solving for A.

Separate glossary example

Simplify (A∩B)∪(A∩Bᶜ), with A,B⊆U: it equals A.

2x+3x=5x

Open glossary card

Step by step

Step 1 / 6

Check the operation’s direction

Difference keeps elements of the first set and removes those in the second. Swapping the arguments changes the meaning.

A∖B=A∩Bᶜ

The complement uses the common universe U.

Worked example

Is A∖B=B∖A true for all sets?

  1. Take A={1,2}, B={2}. Removing 2 from A leaves A∖B={1}.
  2. For B∖A, the only element 2 is removed, yielding ∅.
  3. Since {1}≠∅, the universal law is false. Equality may still hold for other choices of A,B.
Optional self-check

Why is (A∖B)∩B empty?

Show answer

An element would have to be both outside B and inside B, which is impossible.

Source / textbook · approved access required

Open source: printed p. 407 · PDF 431