1240 ch1

All other elements of the template, with equivalent wording being acceptable.

p | ¬p 0 | 1 1 | 0

A biconditional p ↔ q (read 'p if and only if q') is true if and only if (p → q) and (q → p) are both true.

A compound proposition that is false for all truth value assignments of its components.

p(N) is true.

Therefore S(k) → S(k + 1).

By the appropriate Rule of Inference and a reference to the line numbers of the premises.

The Rule of Contradiction states that if ¬p → F0 is true, then p is true. [(¬p → F0) ⇒ p]

A proposition is a declarative sentence that is either true or false.

The disjunction of p and q, p ∨ q, is true if and only if p is true or q is true (or both are true).

F0 = 0, F1 = 1, and for n > 1, Fn = Fn-1 + Fn-2.

Considerations of the axioms of the integers.

Let k ≥ N and suppose S(k) is true.

By the appropriate Law of Logic and a reference to the line number of the earlier statement.

The Rule of Conjunctive Simplification states that if p ∧ q is true, then p is true. [(p ∧ q) ⇒ p]

The conjunction of p and q, p ∧ q, is true if and only if both p and q are true.

S(n0), ..., S(n1) are true because [Justification].

S(n0), S(n0 + 1), ..., S(n1) are true.

[(p → q) ∧ (r → s) ∧ (¬q ∨ ¬s)] ⇒ (¬p ∨ ¬r)

Modus Ponens is a rule of inference where if p and p → q are true, then q is true. [(p ∧ (p → q)) ⇒ q]

0 represents 'false' and 1 represents 'true'.

The implied application of the Law of Universal Generalization: ∀ x ≥ N, S(x) → S(x + 1).

By the principle of mathematical induction, S(n) is true for all integers n ≥ N.

A declarative sentence that contains one or more variables, is not a proposition, and becomes a proposition when variables are replaced with allowable values.

The Rule of Proof by Cases states that if p → r and q → r are true, then (p ∨ q) → r is true. [(p → r) ∧ (q → r)] ⇒ [(p ∨ q) → r]

An implication p → q (read 'p implies q' or 'if p, then q') is true if and only if p being true implies q is true.

p | q | p ∧ q 0 | 0 | 0 0 | 1 | 0 1 | 0 | 0 1 | 1 | 1

Therefore (S(n0) ∧ ... ∧ S(k)) → S(k + 1).

Use definitions and previously proven results to deduce a contradiction.

For all n ≥ n0, S(n) is true.

Numbered lines in two columns: statements on the left and justifications on the right. The last line is the conclusion.

The Law of Syllogism states that if p → q and q → r are true, then p → r is true. [(p → q) ∧ (q → r)] ⇒ (p → r)

A proposition that is not a compound proposition.

It invalidates the inductive step and may result in losing marks.

The assumption that p(k) is true.

[(p → q) ∧ (r → s) ∧ (p ∨ r)] ⇒ (q ∨ s)

p logically implies q, written p ⇒ q, if the proposition p → q is a tautology.

0 0 1, 0 1 0, 1 0 0, 1 1 1

p | q | p ∨ q 0 | 0 | 0 0 | 1 | 1 1 | 0 | 1 1 | 1 | 1

∀ k ≥ N, p(k) → p(k + 1) is true.

The Rule of Conjunction states that if p and q are true, then p ∧ q is true. [(p ∧ q) → (p ∧ q)]

To make the definition unambiguous. For example, (a ∧ b) ∨ c is different from a ∧ (b ∨ c).

Let k ≥ n1 and suppose S(n0), S(n0 + 1), ..., S(k) are true.

Assume p is true and that q is false. Introduce all necessary variables.

For any k ≥ n1, if S(n0), ..., S(k) are all true, then S(k + 1) is also true.

Modus Tollens is a rule of inference where if p → q and ¬q are true, then ¬p is true. [(p → q) ∧ ¬q] ⇒ ¬p

The negation of p, ¬p, is true if and only if the proposition p is false.

Propositions s1 and s2 are logically equivalent (written s1 ⇔ s2) if s1 is true whenever s2 is true, and vice versa.

∀ n ≥ N, p(n) is true.

Universal and existential quantifiers.

p1, p2, ..., pn are the premises.

0 0 1, 0 1 1, 1 0 0, 1 1 1

A compound proposition that is true for all truth value assignments of its components.

By the principle of mathematical induction, S(n) is true for all integers n ≥ n0.

Conclude that p → q is true.

S(N) is true because ... [Justification]

With the justification 'Premise'.

The Rule of Disjunctive Syllogism states that if p ∨ q and ¬p are true, then q is true. [(p ∨ q) ∧ ¬p] ⇒ q

A proposition formed by connecting two propositions with conjunction, disjunction, implication, or biconditional or by the negation of a proposition.

Italicized elements in square parentheses and underlined headings explaining the template.

The step where p(N) is shown to be true.

p ⇒ (p ∨ q)

q is the conclusion.

The step where p(k) is assumed to be true and used to prove p(k + 1).

[(p ∧ q) ∧ [p → (q → r)]] ⇒ r

An argument is valid if the conclusion is true whenever all of the premises are true.