Write a complete argument, then compare it with a worked solution and review checklist. Your written reasoning is not automatically graded.
In x+y=0, each variable keeps its chosen value throughout the statement. If x=3, only y=−3 satisfies this condition.
x+y=0; x=3 ⇒ y=−3
The condition x>0 depends on the permitted values of x. Specify the domain: integers.
x∈ℤ
T / F; ⊤ / ⊥
T and F denote true and false. The symbols ⊤ and ⊥ are truth and falsity constants in the stated logical notation.
In digital logic, 1 and 0 encode these values. They do not make the logical operation OR the same as ordinary integer addition: 1∨1=1.
The statement 3 = 3 has truth value T.
¬ ~
Reverses the truth value of a statement.
Read negation as: it is not the case that…
Epp commonly writes ~p for negation. In a~b the same-looking mark may instead name a relation; read the whole construction.
¬(2 < 3) is false.
∧
True exactly when both statements are true.
One false part makes the conjunction false.
(2<3) ∧ (4<5) is true.
∨
True when at least one statement is true, including when both are true.
Logical OR is inclusive unless stated otherwise.
(2<3) ∨ (4<3) is true.
⊕
True exactly when one of the two statements is true.
Unlike inclusive OR, XOR is false when both parts are true.
T ⊕ T = F
≡
Two formulas have the same truth value for every allowable assignment.
The sign ≡ also has a different use in modular arithmetic.
Logical ≡ is different from a≡b (mod m), which compares integer remainders. State the context before reading the sign.
¬(p ∧ q) ≡ ¬p ∨ ¬q
¬(p∧q)≡¬p∨¬q; ¬(p∨q)≡¬p∧¬q
For classical propositions p and q, negating a conjunction gives a disjunction of the negations, and negating a disjunction gives a conjunction of the negations.
Negate the complete parenthesized expression and each component, including the strictness of inequalities. ≡ says the two formulas agree under every truth assignment.
For real x, ¬(x>0 ∧ x<1) is equivalent to x≤0 ∨ x≥1.
A declarative sentence that is either true or false, but not both.
A question such as “How many?” is not a statement.
The statement 2 < 3 is true.
Lists every truth-value assignment to the input propositions and the resulting values of the formula; intermediate columns can show its parts.
A table with n independent propositional inputs has 2^n rows. T… is not a mathematical symbol for a truth table.
For ¬p: when p=T, ¬p=F; when p=F, ¬p=T.
For p and q: TT, TF, FT, FF.
A formula that is true for every assignment of truth values.
At least one of these two parts is always true.
p ∨ ¬p
A formula that is false for every assignment of truth values.
A statement and its negation cannot both be true.
p ∧ ¬p
Find the value of an expression using the given values of its variables.
The inputs are supplied and the requested result is a value. Solving instead asks which inputs satisfy a stated condition.
Evaluate p∨q when p=F and q=T: the value is T.
Evaluate 2x+1 at x=3: 2×3+1=7.
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.
Simplify (A∩B)∪(A∩Bᶜ), with A,B⊆U: it equals A.
2x+3x=5x
Take arbitrary odd integers m and n.
Reason: Their representations follow from the definition of odd.
This is the goal, not something already known.
Reason: An existential conclusion requires constructing a witness.
Expand before introducing s.
Reason: Substitution and the distributive law.
The constructed integer s proves the product odd.
Reason: Closure of integers and the definition of odd; the goal was not assumed.
When to choose: Choose it when definitions turn the hypothesis into usable equations or properties.
For every x in D: P(x) implies Q(x). Given: an arbitrary x in D satisfying P. Goal: Q for that same x.
Start: Introduce the objects and their domains; assume only the stated hypotheses. Expand the relevant definition.
Finish: State Q and explain why arbitrariness gives the universal claim. Never use Q as a premise.
Take arbitrary positive integers a and b with a dividing b.
Reason: These are the hypotheses, not conclusions.
We must establish this inequality.
Reason: Writing a goal does not establish it.
Introduce an integer q witnessing divisibility.
Reason: Definition of a dividing b; a is nonzero.
The quotient is positive.
Reason: Both b and a are positive; division by a is allowed.
A positive integer cannot lie strictly between 0 and 1.
Reason: This uses q being an integer, not positivity alone.
Replace aq by b to obtain the goal.
Reason: Multiplication by a positive a preserves the inequality; b=aq was established above.
When to choose: Use it to refute a universal claim. A failed proof attempt alone does not refute anything.
To refute every x in D satisfying P also satisfies Q, find one x in D with P true and Q false.
Start: Write the exact claim and choose a permitted test object. Verify its hypotheses.
Finish: Show explicitly which conclusion fails; one valid counterexample refutes the universal claim.
Test the claim that every integer greater than 1 is prime.
Reason: A universal statement can be disproved by one permitted instance.
The chosen value satisfies the domain and hypothesis.
Reason: A counterexample must meet every hypothesis.
Nine has a positive divisor other than 1 and itself, so it is not prime.
Reason: Definition of prime. Composite means nonprime only within integers greater than 1.
When to choose: Choose cases when a definition or remainder changes the calculation. First show the cases cover every allowed object.
Given P, cover all possibilities C₁,…,Cₖ and prove Q in each.
Start: Derive the case split from a definition or theorem; do not merely list convenient examples.
Finish: Conclude Q because every allowed object belongs to a handled case.
Let n be an arbitrary odd integer.
Reason: The hypothesis fixes the domain but does not yet provide the required m.
We need to construct an integer m.
Reason: The conclusion is existential: both the equation and integrality must be shown.
Only r=1 or r=3 is possible for odd n.
Reason: Quotient–remainder theorem 4.5.1; residues 0 and 2 make n even. This proves completeness.
Case 1: set m=2q²+q.
Reason: Expand the square and factor 8. Sums and products of integers are integers.
Case 2: set m=2q²+3q+1.
Reason: Expansion, 9=8+1, then factor 8; the constructed m is an integer.
Both possible cases give the required integer; it may depend on the case.
Reason: Exhaustive cases establish the original claim for arbitrary odd n.
When to choose: Use it when negating the conclusion gives a more useful starting definition.
To prove P⇒Q, prove ¬Q⇒¬P. Given ¬Q; goal ¬P.
Start: Write the contrapositive and introduce an arbitrary object satisfying ¬Q.
Finish: Establish ¬P, then invoke equivalence with the original implication.
Prove the statement for every integer n.
Reason: This is Proposition 4.7.4.
Instead prove: if n is odd, then n² is odd.
Reason: P⇒Q is equivalent to ¬Q⇒¬P. For integers, not even means odd. We do not assume n² even here.
Introduce an integer k.
Reason: Definition of odd.
The square is odd.
Reason: Algebra and closure of integers show 2k²+2k is an integer; then use the definition of odd.
The original implication follows from the proved contrapositive.
Reason: Logical equivalence of a conditional and its contrapositive.
When to choose: Use it when the negation creates incompatible established facts.
Assume the hypotheses and the negation of the desired conclusion; derive a contradiction.
Start: State exactly what is temporarily assumed for contradiction.
Finish: Name the conflicting statements and discharge the assumption.
Prove the statement for every integer n.
Reason: This is Proposition 4.7.4.
Assume a counterexample: n² is even but n is not even.
Reason: Negating the universal conditional gives one integer satisfying the hypothesis and negating the conclusion. Every integer is either even or odd.
Introduce an integer k.
Reason: Definition of odd.
The square is odd.
Reason: Algebra and closure of integers show 2k²+2k is an integer; then use the definition of odd.
This is impossible; the assumed counterexample cannot exist.
Reason: No integer is both even and odd (Theorem 4.7.2). Discharging the contradictory assumption proves the original claim.
When to choose: Use it for a claim indexed by every integer n from a starting value, when the next case relates to the previous one.
Prove P(n₀). For arbitrary k≥n₀, assume P(k) and derive P(k+1).
Start: State the indexed property and starting value; check the base without using the induction hypothesis.
Finish: Invoke induction only after both the base and the implication are established.
Prove the sum formula for every positive integer n.
Reason: The domain and indexed property specify what induction must cover.
The first case holds.
Reason: Direct evaluation establishes the base.
Fix an arbitrary integer k and assume only its case.
Reason: This temporary induction hypothesis is permitted to prove the implication P(k)⇒P(k+1).
Separate the next term and replace the old sum by k².
Reason: This is the exact point where the induction hypothesis is used.
This is the required case k+1.
Reason: Algebra proves the implication. The base and induction principle now give all integers n≥1.
When to choose: Use it to identify precisely what would make a quantified claim false, especially before a counterexample or contradiction.
Negate one outer quantifier at a time, preserving variable order, domains and the scope of the predicate.
Start: Mark the scope of each quantifier; move the negation inward one rule at a time.
Finish: Read the final statement and check that it describes failure of the original claim. Equivalence does not by itself prove either statement true.
Rewrite the negation without changing its meaning.
Reason: The domains D and E remain fixed throughout.
There is an x for which the inner existential statement fails.
Reason: Negation of a universal quantifier: ¬∀x R(x) ⇔ ∃x ¬R(x).
For that same x, every y fails P.
Reason: Negation of an existential quantifier: ¬∃y P(x,y) ⇔ ∀y ¬P(x,y). The order ∃x∀y is preserved.
Both are statements established by proof. The name often signals how a text organizes or emphasizes a result, not a different degree of truth.
A proved auxiliary result used in another argument. For example, −|r|≤r≤|r| for real r supports later bounds. If r≥0, |r|=r and −r≤r; if r<0, |r|=−r and r≤−r. These cases cover every real r.
Lemma 4.5.4, printed p. 207 / PDF 231
A result derived from an established theorem. If r is rational, r+r is rational by closure of rational numbers under addition (Theorem 4.3.2); because 2r=r+r, its double is rational.
Example 4.3.4, printed p. 187 / PDF 211; printed box: Corollary 4.2.3
With Epp’s definition, d divides 0 for every nonzero integer d: 0=d·0. Do not omit d≠0.
Prime and composite classifications here concern integers n>1. “Not prime” alone does not make 0, 1 or a negative integer composite.
Check answer checks the result above. Your reasoning is saved, not automatically graded; compare it with the walkthrough.
Back to §2.1 in All topicsConcept in Epp: §2.1, printed page 37 ↗
¬p ∧ ¬q requires both parts to be false and is too strong.