1.2 Propositional Logic and Quantifiers

1.2 Propositional Logic and Quantifiers

Mathematical reasoning relies on rigorous logical structures. On the Gaokao, logic problems test your ability to evaluate the truth values of mathematical assertions, determine necessary and sufficient conditions, negate quantified statements, and execute contrapositive proofs. Mastery of logical connectives and quantifier negation ensures precise problem analysis across analysis, geometry, and algebra.

1. Propositions and Compound Logical Statements

A proposition (命题) is a declarative statement that is either definitively true (T) or definitively false (F), but not both. Truth or falsity is called the truth value (真假值) of the proposition.

Simple and Compound Propositions

• Simple Proposition: A single assertion containing no logical operators (e.g., "$2$ is an even prime number").
• Compound Proposition: Formed by combining simple propositions using logical connectives: and (且, $\land$), or (或, $\lor$), and not (非, $\neg$).

Truth Rules for Compound Propositions

1. Conjunction ($p \land q$): True if and only if both $p$ and $q$ are true. If either is false, $p \land q$ is false.
2. Disjunction ($p \lor q$): True if at least one of $p$ or $q$ is true. False only when both $p$ and $q$ are false.
3. Negation ($\neg p$): Reverses the truth value of $p$. If $p$ is true, $\neg p$ is false; if $p$ is false, $\neg p$ is true.

2. Sufficient, Necessary, and Necessary-and-Sufficient Conditions

In mathematical statements of the form "If $p$, then $q$" ($p \implies q$):

Definitions

1. Sufficient Condition (充分条件): If $p \implies q$ is true, then $p$ is a sufficient condition for $q$.
2. Necessary Condition (必要条件): If $p \implies q$ is true, then $q$ is a necessary condition for $p$.
3. Necessary and Sufficient Condition (充要条件): If both $p \implies q$ and $q \implies p$ are true ($p \iff q$), then $p$ is a necessary and sufficient condition (equivalent condition) for $q$.

Set-Theoretic Interpretation (小范围推大范围)

Let $A = {x \mid p(x)}$ be the solution set where condition $p$ holds, and $B = {x \mid q(x)}$ be the solution set where condition $q$ holds:

• If $A \subseteq B$, then $p \implies q$ (so $p$ is sufficient for $q$).
• If $B \subseteq A$, then $q \implies p$ (so $p$ is necessary for $q$).
• If $A = B$, then $p \iff q$ (so $p$ is necessary and sufficient for $q$).
• If $A \subsetneq B$, then $p$ is a sufficient but not necessary condition for $q$.
• If $B \subsetneq A$, then $p$ is a necessary but not sufficient condition for $q$.
• If neither $A \subseteq B$ nor $B \subseteq A$, then $p$ is neither sufficient nor necessary for $q$.

3. Four Forms of Propositions and Contrapositive Equivalence

For a conditional proposition "If $p$, then $q$":

• Original Proposition (原命题): If $p$, then $q$ ($p \implies q$).
• Converse Proposition (逆命题): If $q$, then $p$ ($q \implies p$).
• Inverse Proposition (否命题): If $\neg p$, then $\neg q$ ($\neg p \implies \neg q$).
• Contrapositive Proposition (逆否命题): If $\neg q$, then $\neg p$ ($\neg q \implies \neg p$).

Logical Equivalence Principle

• The Original Proposition and its Contrapositive are logically equivalent: $(p \implies q) \equiv (\neg q \implies \neg p)$.
• The Converse Proposition and Inverse Proposition are logically equivalent: $(q \implies p) \equiv (\neg p \implies \neg q)$.
• Gaokao Technique: When proving a statement $p \implies q$ directly is difficult, prove its contrapositive $\neg q \implies \neg p$ instead.

4. Universal and Existential Quantifiers and Their Negations

Definitions

• Universal Quantifier (全称量词): "For all", "For every", denoted $\forall$.
  • Universal Proposition: $\forall x \in M, p(x)$.
• Existential Quantifier (存在量词): "There exists", "At least one", denoted $\exists$.
  • Existential Proposition: $\exists x \in M, p(x)$.

Negation Rules for Quantified Propositions

To negate a quantified proposition, perform two simultaneous transformations: swap the quantifier symbol and negate the predicate property, keeping the domain unchanged:

1. $\neg \left( \forall x \in M, p(x) \right) \equiv \exists x \in M, \neg p(x)$
2. $\neg \left( \exists x \in M, p(x) \right) \equiv \forall x \in M, \neg p(x)$

Summary of Quantifier Forms, Negations, and Logical Equivalences

5. Worked Gaokao Exam Examples

Worked Example 1: Necessary and Sufficient Condition Analysis

Problem: Determine whether condition $p: m > 2$ is a sufficient, necessary, or充要 (necessary and sufficient) condition for condition $q$: the quadratic equation $x^2 + 2x + m = 0$ has no real roots.

Solution:

• Step 1: Simplify condition $q$. The quadratic equation $x^2 + 2x + m = 0$ has no real roots if and only if its discriminant $\Delta < 0$: $\Delta = b^2 - 4ac = 2^2 - 4(1)(m) = 4 - 4m < 0$ Solving $4 - 4m < 0 \implies 4m > 4 \implies m > 1$. Thus, condition $q$ is equivalent to $m > 1$.
• Step 2: Compare the solution sets of $p$ and $q$. Let $A = {m \mid m > 2} = (2, +\infty)$ be the set for $p$. Let $B = {m \mid m > 1} = (1, +\infty)$ be the set for $q$.
• Step 3: Analyze set inclusion. Notice that $(2, +\infty) \subsetneq (1, +\infty)$, which means $A \subsetneq B$.
  • Since $A \subseteq B$, $p \implies q$ is true (if $m > 2$, then $m > 1$ is guaranteed).
  • Since $B \not\subseteq A$, $q \implies p$ is false (for example, if $m = 1.5$, $q$ holds but $p$ fails).
• Conclusion: Condition $p$ ($m > 2$) is a sufficient but not necessary condition for condition $q$.

Worked Example 2: Quantifier Negation and Parameter Separation

Problem: Given that the proposition $p: \forall x \in [1, 2], x^2 - ax + 1 \ge 0$ is true, find the range of real values of the parameter $a$.

Solution:

• Step 1: Reformulate the universal proposition using parameter separation. The inequality $x^2 - ax + 1 \ge 0$ holds for all $x \in [1, 2]$. Since $x \in [1, 2]$, we know $x > 0$. We can isolate $a$ by adding $ax$ to both sides and dividing by $x$: $ax \le x^2 + 1 \implies a \le \frac{x^2 + 1}{x} = x + \frac{1}{x}$
• Step 2: Apply the maximum/minimum principle for universal conditions. For $a \le f(x)$ to hold for all $x \in [1, 2]$, $a$ must be less than or equal to the minimum value of $f(x) = x + \frac{1}{x}$ on the interval $[1, 2]$: $a \le \min_{x \in [1, 2]} \left( x + \frac{1}{x} \right)$
• Step 3: Find the minimum of $f(x) = x + \frac{1}{x}$ on $[1, 2]$. By the AM-GM inequality, for $x > 0$: $x + \frac{1}{x} \ge 2 \sqrt{x \cdot \frac{1}{x}} = 2$ Equality holds when $x = \frac{1}{x} \implies x^2 = 1 \implies x = 1$. Since $x = 1 \in [1, 2]$, the minimum value on $[1, 2]$ is indeed $2$.
• Step 4: State the final parameter range. $a \le 2 \implies a \in (-\infty, 2]$