Lecture 1: What is a Proof?
Propositional Logic Foundations
Def 1Statement
A declarative sentence that is either true or false, but not both.
Def 2Proposition
A statement whose truth value is well-defined and can be determined to be true or
false.
Def 3Truth Value
One of the two logical constants: True (T, 1) or False (F, 0).
Def 4Propositional Variable
A variable (typically denoted p, q, r, etc.) that represents a proposition and can
take truth values T or F.
Def 5Atomic Proposition
A simple, indivisible proposition that cannot be broken down into simpler
propositions.
Def 6Compound Proposition
A proposition formed by combining atomic propositions using logical connectives.
Def 7Logical Connectives
Given propositions p and q:
- Negation: \( \neg p \) (NOT p) — opposite truth value
- Conjunction: \( p \land q \) (p AND q) — true only if both are true
- Disjunction: \( p \lor q \) (p OR q) — true if at least one is true
- Exclusive OR: \( p \oplus q \) (p XOR q) — true if exactly one is true
- Implication: \( p \to q \) (if p then q) — false only when p is true and q is false
- Biconditional: \( p \leftrightarrow q \) (p if and only if q) — true when both have same truth value
Def 8Well-Formed Formula (WFF)
A finite expression constructed from propositions using logical connectives
according to syntactic rules. Every WFF is syntactically correct.
Def 9Truth Assignment
A function assigning each propositional variable a value from \( \{T, F\} \). Also
called a valuation or interpretation.
Def 10Truth Table
A table showing all possible truth values of a formula for all possible truth
assignments of its variables.
Def 11Semantics
The meaning or truth value of a formula under a given truth assignment.
Formula Classifications
Def 12Tautology
A formula that evaluates to true under every truth assignment. Example: \( p \lor
\neg p \) (Law of Excluded Middle)
Def 13Contradiction
A formula that evaluates to false under every truth assignment. Example: \( p \land
\neg p \) (Law of Non-Contradiction)
Def 14Contingency
A formula that is neither a tautology nor a contradiction; it is true for some
truth assignments and false for others.
Def 15Satisfiable Formula
A formula that evaluates to true under at least one truth assignment. Both
tautologies and contingencies are satisfiable.
Def 16Unsatisfiable Formula
A formula that evaluates to false under all truth assignments (equivalent to a
contradiction).
Def 17Logical Equivalence
Two formulas \( \phi \) and \( \psi \) are logically equivalent (denoted \( \phi
\equiv \psi \) or \( \phi \Leftrightarrow \psi \)) if they have the same truth value under all truth
assignments.
Def 18Logical Implication
A formula \( \phi \) logically implies \( \psi \) (denoted \( \phi \models \psi \))
if whenever \( \phi \) is true, \( \psi \) must also be true.
Arguments and Reasoning
Def 19Argument
A finite sequence of propositions consisting of premises \( p_1, p_2, \ldots, p_n
\) followed by a conclusion \( q \).
Def 20Valid Argument
An argument where, whenever all premises are true, the conclusion must also be
true. Formally: \( (p_1 \land p_2 \land \cdots \land p_n) \to q \) is a tautology.
Def 21Sound Argument
A valid argument where all premises are actually true. Sound arguments always have
true conclusions.
Def 22Fallacy
An invalid form of reasoning or argument that appears correct but contains a
logical error.
Def 23Inference Rule
A logical pattern that allows deriving a conclusion from premises. Examples: Modus
Ponens, Modus Tollens, Hypothetical Syllogism.
Predicate Logic & Quantifiers
Def 24Predicate
A statement involving variables that becomes a proposition when specific values are
assigned to the variables. Example: P(x) = "x is even"
Def 25Domain of Discourse (Universe)
The set of all possible values that variables in a predicate can take. Denoted as \(
\mathcal{U} \) or \( D \).
Def 26Universal Quantifier
The symbol \( \forall \) (for all) that asserts a predicate is true for every element in the domain.
Notation: \( \forall x \, P(x) \) means "P(x) is true for all x in the domain"
Notation: \( \forall x \, P(x) \) means "P(x) is true for all x in the domain"
Def 27Existential Quantifier
The symbol \( \exists \) (there exists) that asserts a predicate is true for at least one element in the
domain.
Notation: \( \exists x \, P(x) \) means "there exists at least one x such that P(x) is true"
Notation: \( \exists x \, P(x) \) means "there exists at least one x such that P(x) is true"
Def 28Unique Existential Quantifier
The symbol \( \exists! \) that asserts a predicate is true for exactly one element in the domain.
Notation: \( \exists! x \, P(x) \) means "there exists exactly one x such that P(x) is true"
Notation: \( \exists! x \, P(x) \) means "there exists exactly one x such that P(x) is true"
Def 29Bound Variable
A variable that appears within the scope of a quantifier. Example: in \( \forall x \,
P(x) \), x is bound.
Def 30Free Variable
A variable that is not bound by any quantifier. Example: in \( P(x) \land Q(y, x) \),
if only x is quantified, y is free.
Def 31Nested Quantifiers
Multiple quantifiers appearing in sequence. The order matters!
- \( \forall x \, \forall y \, P(x,y) \) — for all x and all y, P(x,y) is true
- \( \exists x \, \exists y \, P(x,y) \) — there exist x and y such that P(x,y) is true
- \( \forall x \, \exists y \, P(x,y) \) — for every x, there exists a y such that P(x,y)
- \( \exists x \, \forall y \, P(x,y) \) — there exists an x such that for all y, P(x,y)
Important
Negation of Quantifiers:
- \( \neg (\forall x \, P(x)) \equiv \exists x \, \neg P(x) \)
- \( \neg (\exists x \, P(x)) \equiv \forall x \, \neg P(x) \)
Examples of Predicates
Ex 1Simple Predicate
Let P(x) = "x > 5" with domain \( \mathbb{N} \). Then P(7) is true, but P(3) is
false.
Ex 2Universal Statement
\( \forall x \in \mathbb{Z} \, (x^2 \geq 0) \) — "The square of every integer is
non-negative" (TRUE)
Ex 3Existential Statement
\( \exists x \in \mathbb{R} \, (x^2 = 2) \) — "There exists a real number whose
square is 2" (TRUE, x = √2)
Ex 4Nested Quantifiers
\( \forall x \in \mathbb{R} \, \exists y \in \mathbb{R} \, (y > x) \) — "For
every real number, there exists a larger one" (TRUE)
Proof Techniques
Def 32Proof
A finite sequence of logically valid steps deriving a conclusion from assumptions
using established rules of inference and axioms.
Def 33Axiom (Postulate)
A statement accepted as true without proof, serving as a starting point for further
reasoning.
Def 34Theorem
A mathematical statement that has been proven to be true based on axioms and
previously proven theorems.
Def 35Lemma
A minor theorem used as a stepping stone to prove a larger theorem. Often called a
"helper theorem."
Def 36Corollary
A statement that follows readily from a theorem with little or no additional proof.
Def 37Conjecture
A statement believed to be true but not yet proven. It remains a conjecture until
proven or disproven.
Proof Methods
Def 38Direct Proof
A proof that derives the conclusion directly from the premises using logical deductions.
Strategy: To prove \( p \to q \), assume p is true and show q must be true through logical steps.
Strategy: To prove \( p \to q \), assume p is true and show q must be true through logical steps.
Def 39Proof by Contrapositive
A proof that establishes an implication \( p \to q \) by proving its
Def 40Proof by Contradiction
A method where one assumes the statement to be proven is false, and then derives a logical contradiction
(e.g., \( r \land \neg r \)). Since contradictions cannot exist, the assumption must be false, making the
original statement true.
Strategy: To prove \( p \), assume \( \neg p \) is true and derive a contradiction.
Strategy: To prove \( p \), assume \( \neg p \) is true and derive a contradiction.
Def 41Proof by Cases
A method used when the hypothesis contains a disjunction of cases. One must prove the conclusion is true for
every possible case.
Strategy: To prove \( (p_1 \lor p_2 \lor \dots \lor p_n) \to q \), prove \( p_1 \to q \), \( p_2 \to q \), ..., and \( p_n \to q \).
Strategy: To prove \( (p_1 \lor p_2 \lor \dots \lor p_n) \to q \), prove \( p_1 \to q \), \( p_2 \to q \), ..., and \( p_n \to q \).
Def 42Mathematical Induction
A technique to prove a statement \( P(n) \) for all natural numbers \( n \geq n_0 \).
- Base Case: Verify \( P(n_0) \) is true.
- Inductive Hypothesis: Assume \( P(k) \) is true for an arbitrary \( k \geq n_0 \).
- Inductive Step: Prove that \( P(k) \to P(k+1) \) is true.
Def 43Existence Proofs
Proofs used to show \( \exists x P(x) \).
- Constructive: Find a specific element \( a \) such that \( P(a) \) is true.
- Non-constructive: Prove that such an element must exist without explicitly finding it (often via contradiction).
Def 44Counterexample
A specific instance that shows a universal statement \( \forall x P(x) \) is false. Finding one
counterexample is sufficient to disprove a universal claim.
Logical Laws & Equivalences
The symbol \( \equiv \) denotes logical equivalence. These laws allow us to simplify complex compound propositions.
| Name of Law | Equivalence |
|---|---|
| Identity Laws | \( p \land T \equiv p \) \( p \lor F \equiv p \) |
| Domination Laws | \( p \lor T \equiv T \) \( p \land F \equiv F \) |
| Idempotent Laws | \( p \lor p \equiv p \) \( p \land p \equiv p \) |
| Double Negation | \( \neg(\neg p) \equiv p \) |
| Commutative Laws | \( p \lor q \equiv q \lor p \) \( p \land q \equiv q \land p \) |
| Associative Laws | \( (p \lor q) \lor r \equiv p \lor (q \lor r) \) \( (p \land q) \land r \equiv p \land (q \land r) \) |
| Distributive Laws | \( p \lor (q \land r) \equiv (p \lor q) \land (p \lor r) \) \( p \land (q \lor r) \equiv (p \land q) \lor (p \land r) \) |
| De Morgan's Laws | \( \neg(p \land q) \equiv \neg p \lor \neg q \) \( \neg(p \lor q) \equiv \neg p \land \neg q \) |
| Absorption Laws | \( p \lor (p \land q) \equiv p \) \( p \land (p \lor q) \equiv p \) |
| Implication Law | \( p \to q \equiv \neg p \lor q \) |
Proof Examples
Ex 5Direct Proof
Theorem: If \( n \) is an odd integer, then \( n^2 \) is odd.
Proof: 1. Assume \( n \) is odd. By definition, \( n = 2k + 1 \) for some integer \( k \).
2. Square both sides: \( n^2 = (2k+1)^2 = 4k^2 + 4k + 1 \).
3. Factor out 2: \( n^2 = 2(2k^2 + 2k) + 1 \).
4. Let \( m = 2k^2 + 2k \). Since \( k \) is an integer, \( m \) is an integer.
5. \( n^2 = 2m + 1 \), which fits the definition of an odd number. Q.E.D.
Proof: 1. Assume \( n \) is odd. By definition, \( n = 2k + 1 \) for some integer \( k \).
2. Square both sides: \( n^2 = (2k+1)^2 = 4k^2 + 4k + 1 \).
3. Factor out 2: \( n^2 = 2(2k^2 + 2k) + 1 \).
4. Let \( m = 2k^2 + 2k \). Since \( k \) is an integer, \( m \) is an integer.
5. \( n^2 = 2m + 1 \), which fits the definition of an odd number. Q.E.D.
Ex 6Proof by Contrapositive
Theorem: If \( 3n + 2 \) is odd, then \( n \) is odd.
Proof: We prove the contrapositive: "If \( n \) is even, then \( 3n + 2 \) is even."
1. Assume \( n \) is even. So \( n = 2k \).
2. Then \( 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) \).
3. Since \( 3k+1 \) is an integer, \( 3n + 2 \) is essentially \( 2 \times \text{integer} \), which is even.
4. Since the contrapositive is true, the original statement is true.
Proof: We prove the contrapositive: "If \( n \) is even, then \( 3n + 2 \) is even."
1. Assume \( n \) is even. So \( n = 2k \).
2. Then \( 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) \).
3. Since \( 3k+1 \) is an integer, \( 3n + 2 \) is essentially \( 2 \times \text{integer} \), which is even.
4. Since the contrapositive is true, the original statement is true.
Ex 7Proof by Contradiction
Theorem: \( \sqrt{2} \) is irrational.
Proof: 1. Assume for the sake of contradiction that \( \sqrt{2} \) is rational.
2. Then \( \sqrt{2} = a/b \) for integers \( a, b \) with no common factors (simplified).
3. \( 2 = a^2/b^2 \implies a^2 = 2b^2 \). Thus \( a^2 \) is even, implying \( a \) is even (\( a=2k \)).
4. Substitute back: \( (2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies 2k^2 = b^2 \).
5. This implies \( b^2 \) is even, so \( b \) is even.
6. Contradiction: We assumed \( a \) and \( b \) had no common factors, but both are even.
7. Therefore, \( \sqrt{2} \) must be irrational.
Proof: 1. Assume for the sake of contradiction that \( \sqrt{2} \) is rational.
2. Then \( \sqrt{2} = a/b \) for integers \( a, b \) with no common factors (simplified).
3. \( 2 = a^2/b^2 \implies a^2 = 2b^2 \). Thus \( a^2 \) is even, implying \( a \) is even (\( a=2k \)).
4. Substitute back: \( (2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies 2k^2 = b^2 \).
5. This implies \( b^2 \) is even, so \( b \) is even.
6. Contradiction: We assumed \( a \) and \( b \) had no common factors, but both are even.
7. Therefore, \( \sqrt{2} \) must be irrational.
Ex 8Counterexample
Conjecture: For every integer \( n \), \( n^2 \geq n \).
Check:
If \( n = 2, 4 \geq 2 \) (True).
If \( n = -3, 9 \geq -3 \) (True).
If \( n = 0, 0 \geq 0 \) (True).
Wait: Is the domain integers or real numbers? If domain is \( \mathbb{R} \), let \( n = 0.5 \).
\( 0.5^2 = 0.25 \). Is \( 0.25 \geq 0.5 \)? False.
\( n = 0.5 \) is a counterexample for the domain of real numbers.
Check:
If \( n = 2, 4 \geq 2 \) (True).
If \( n = -3, 9 \geq -3 \) (True).
If \( n = 0, 0 \geq 0 \) (True).
Wait: Is the domain integers or real numbers? If domain is \( \mathbb{R} \), let \( n = 0.5 \).
\( 0.5^2 = 0.25 \). Is \( 0.25 \geq 0.5 \)? False.
\( n = 0.5 \) is a counterexample for the domain of real numbers.
Quick Reference
📋 Foundations
Propositions are statements that are strictly True or False. We build complex logic using connectives (\( \land, \lor, \neg, \to \)).
🔍 Quantifiers
Universal (\( \forall \)): True for all elements.
Existential (\( \exists
\)): True for at least one element.
Remember: \( \neg(\forall x P(x)) \equiv \exists x \neg
P(x) \).
🛠️ Proof Strategy
1. Try Direct Proof first.
2. If "If...then", try Contrapositive.
3. If proving non-existence or irrationality, try Contradiction.
4. If dealing with \( \mathbb{N} \), try Induction.
⚠️ Common Pitfalls
- Confusing \( p \to q \) with \( q \to p \) (Converse error).
- Assuming "some" implies "not all".
- Division by zero in algebraic steps.
- Circular reasoning (assuming what you want to prove).
Review
Key Equivalences to Memorize:
- \( p \to q \equiv \neg p \lor q \)
- \( p \to q \equiv \neg q \to \neg p \)
- \( \neg (p \to q) \equiv p \land \neg q \) (Negation of Implication)
- \( p \leftrightarrow q \equiv (p \to q) \land (q \to p) \)