Lukas' Notes

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2026-10-15

3 min read

Chapter 2: Propositional Logic

  • Evaluation of propositional formulas
    • The map takes a formula and an assignment and returns its truth value .
    • The assignment gives truth values to variables; its extension evaluates entire formulas.
  • Disjunctive normal form (DNF)
    • A formula is in DNF if it is a disjunction of conjunctions of literals, where a literal is a variable or a negated variable:

Misplaced & \bigvee_{i=1}^{n}\bigwedge_{j=1}^{m_i}L_{ij}. <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

- Finite disjunctions abbreviate repeated use of $\lor$. The one-term and empty cases are included:

Misplaced & \begin{aligned} \bigvee_{i=1}^{n}A_i&=A_1\lor\cdots\lor A_n &&(n\geq1),\\ \bigvee_{i=1}^{1}A_i&=A_1,\\ \bigvee_{i=1}^{0}A_i&=\bot. \end{aligned} <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

- Dually, the empty conjunction is true: $\bigwedge_{i=1}^{0}A_i=\top$. A disjunction needs a true term; an empty conjunction has no condition that could fail.
  • Existence of an equivalent DNF
    • For every formula , there is a formula in DNF with the same meaning: , i.e., for every assignment .
    • Construction by truth table: let be the variables occurring in . For each row with , form a conjunction containing if and if .
      • This conjunction is true exactly for that row. Disjoining these conjunctions therefore reproduces precisely the rows where is true.
      • If no row makes true, the disjunction is empty and equals .
  • Conjunctive normal form (CNF; German: KNF)
    • A formula is in CNF if it is a conjunction of disjunctions of literals:

Misplaced & \bigwedge_{i=1}^{n}\bigvee_{j=1}^{m_i}L_{ij}. <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

  • Tautology and contradiction
    • A formula is a tautology exactly when : every assignment makes it true.
    • A formula is a contradiction exactly when , equivalently : every assignment makes it false.
    • These are statements about the resulting truth values under all assignments, not about the syntactic shape of the formulas.

Set Theory

Misplaced & (x,y):=\bigl\{\{x\},\{x,y\}\bigr\}. <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

- The representation must preserve the two positions: $(x,y)=(a,b)$ implies $x=a$ and $y=b$. The proof uses unions and intersections to recover the components.
  • Union of a set of sets
    • For any set , define

Misplaced & z\in\bigcup\mathcal{A} \quad\Longleftrightarrow\quad \exists B\in\mathcal{A}:\ z\in B. <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

- The union collects the elements of the members of $\mathcal{A}$, removing one layer of set membership. In set theory, those members are themselves sets.
- For $\mathcal{A}=\{\mathbb{N}\}$, we have $\bigcup\mathcal{A}=\mathbb{N}$. In particular, $7\in\mathbb{N}\in\mathcal{A}$, but $7\notin\{\mathbb{N}\}$: membership is not transitive.
  • Recovering the components of a Kuratowski pair

Misplaced & \begin{aligned} \bigcap p &=\{z\mid\forall C\in p:\ z\in C\} =\{x\}=\{a\},\\ \bigcup p&=\{x,y\}=\{a,b\}. \end{aligned} <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

- Taking the union of the singleton intersection recovers the first component: $\bigcup(\bigcap p)=x=a$.
- To recover the second component, distinguish two cases:
	- If $y=x$, then $\{x,y\}=\{x\}=\{x,b\}$, so $b=x=y$.
	- If $y\neq x$, then $y\in\{x,b\}$ forces $y=b$.
- Thus equality of the representing sets gives equality in each position. Conversely, $x=a$ and $y=b$ immediately give $(x,y)=(a,b)$.
  • Tuples
    • A pair is a -tuple. To represent a triple, we likewise require

Misplaced & (x,y,z)=(a,b,c) \quad\Longleftrightarrow\quad x=a\ \land\ y=b\ \land\ z=c. <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

- One convention is to nest ordered pairs: $(x,y,z):=((x,y),z)$. Applying the pair property twice recovers all three positions.
- Another is to tag each position:

Misplaced & (x,y,z):=\{(0,x),(1,y),(2,z)\}. <span class="math-indent" aria-hidden="true">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span>

	- This is the graph of a [[Knowledge/Function|function]] $\{0,1,2\}\to\{x,y,z\}$. The distinct indices preserve the positions even when some component values coincide.