2.3 Quantifiers & Formal Logic Sets

2.3 Quantifiers & Formal Logic Sets

While strict conditional logic deals in absolute binaries (Yes/No, 100%/0%), real-world arguments frequently involve degrees of scope. The LSAT tests these degrees using quantifiers: words like all, most, many, some, few, several, and none. Interpreting quantifier ranges precisely and knowing which quantifier statements can be combined is essential for mastering formal logic questions.

---

Understanding Quantifiers in LSAT Formal Logic

On the LSAT, quantifiers have strict, standardized definitions that often differ from everyday conversational English.

The Quantifier Spectrum

0%                  1%                              51%                   100%
├───────────────────┼────────────────────────────────┼──────────────────────┤
│   NONE (0%)       │         SOME (1% - 100%)       │   MOST (51% - 100%)  │
│                   │   (At least one / Possibility)  │   (Majority / >50%)  │
└───────────────────┴────────────────────────────────┴──────────────────────┘
                                                     │       ALL (100%)     │
                                                     └──────────────────────┘

Critical Quantifier Definitions

1. SOME means "At least one": In everyday speech, if someone says "Some of my friends are lawyers," they usually imply "not all." On the LSAT, SOME includes the possibility of ALL. "Some A are B" simply guarantees that the intersection of Set A and Set B contains at least one item.
2. MOST means "More than half": MOST ranges from 51% to 100%. It does not mean "almost all" or "the vast majority." 50.1% satisfies MOST.
3. MANY / SEVERAL / FEW:
   • Many and Several are categorized as SOME (at least one, range unknown).
   • Few means "Some are, but Most are not."

---

The Properties & Symmetry of Quantifiers

Understanding symmetry allows you to instantly manipulate statements during analysis.

Symmetric (Reversible) Quantifiers

• SOME: If "Some archaeologists ($A$) are scuba divers ($S$)", it is 100% logically equivalent to say "Some scuba divers ($S$) are archaeologists ($A$)". $A \xleftrightarrow{\text{some}} S \quad \iff \quad S \xleftrightarrow{\text{some}} A$
• NONE: If "No reptiles ($R$) have feathers ($F$)", it is logically equivalent to say "No feathered creatures ($F$) are reptiles ($R$)". $R \longrightarrow \neg F \quad \iff \quad F \longrightarrow \neg R$

Asymmetric (Non-Reversible) Quantifiers

• ALL: Knowing "All golden retrievers ($G$) are dogs ($D$)" does NOT allow you to infer "All dogs are golden retrievers." (That would be a Mistaken Reversal).
• MOST: Knowing "Most billionaires ($B$) are over 40 years old ($O$)" does NOT allow you to infer "Most people over 40 years old are billionaires."

---

Formal Logic Inference Rules & Invalidation Controls

When multiple quantifier statements are combined in a stimulus, specific mathematical rules dictate whether a valid inference can be drawn.

Rule 1: The Left-Hand "ALL" Bridge (Universal Combination)

You can combine a quantifier statement with a universal conditional statement ONLY IF the universal statement points away from the common term or serves as the left-hand base.

$\text{Premise 1: } A \xrightarrow{\text{most}} B$ $\text{Premise 2: } B \longrightarrow C$ $\text{Valid Inference: } A \xrightarrow{\text{most}} C$

Logic: If most A's fall into bucket B, and every single item in bucket B automatically moves to bucket C, then most A's must end up in bucket C.

Rule 2: The Double-Most Overlap Principle

If two MOST statements originate from the SAME left-hand base, their majorities must overlap, yielding a valid SOME inference.

$\text{Premise 1: } A \xrightarrow{\text{most}} B \quad ( >50\% \text{ of } A)$ $\text{Premise 2: } A \xrightarrow{\text{most}} C \quad ( >50\% \text{ of } A)$ $\text{Valid Inference: } B \xleftrightarrow{\text{some}} C$

Mathematical Proof: Imagine Set A has 100 members. Most A are B means at least 51 are B. Most A are C means at least 51 are C. Even if you separate B and C as much as possible ($51 + 51 = 102$), at least 2 members must belong to both B and C. Thus, at least one B must be a C.

Invalid Quantifier Combinations (NO INFERENCE ALLOWED)

---

Overlap Logic, Set Theory & Venn Diagrams on the LSAT

When LSAT problems involve numerical overlaps or set intersections, visualizing the relationships via overlapping set tables or Venn diagrams prevents errors.

+-----------------------------------------------------------------------+
|                         SET OVERLAP MATRIX                            |
|                                                                       |
|   Total Group A Population = 100                                      |
|   ├── Members in Set B (Most)  = 60                                   |
|   └── Members in Set C (Most)  = 70                                   |
|                                                                       |
|   Minimum Overlap (B ∩ C) = (60 + 70) - 100 = 30                      |
|   Maximum Overlap (B ∩ C) = min(60, 70)     = 60                      |
|                                                                       |
|   Conclusion: At least 30 members (SOME) belong to both B and C.       |
+-----------------------------------------------------------------------+

---

Concrete LSAT Worked Examples & Deep Strategic Analysis

Worked Example 1: Multi-Quantifier Inference Chain

Passage: All senior partners at Law Firm X are equity shareholders. Most senior partners at Law Firm X specialize in corporate litigation. No attorney who specializes in corporate litigation works on pro bono publico cases.

Step 1: Formalize each statement:

1. Senior Partner ($SP$) $\longrightarrow$ Equity Shareholder ($ES$).
2. Senior Partner ($SP$) $\xrightarrow{\text{most}}$ Corporate Litigator ($CL$).
3. Corporate Litigator ($CL$) $\longrightarrow \neg$ Pro Bono ($PB$).

Step 2: Derive Valid Inferences:

• Inference A (Combining 2 & 3): Since $SP \xrightarrow{\text{most}} CL$ and $CL \rightarrow \neg PB$, we get: $SP \xrightarrow{\text{most}} \neg PB$ ("Most senior partners do not work on pro bono cases").
• Inference B (Double Base on $SP$): $SP \rightarrow ES$ means All $SP$ are $ES$. Combined with $SP \xrightarrow{\text{most}} CL$, we get: $ES \xleftrightarrow{\text{some}} CL$ ("Some equity shareholders specialize in corporate litigation").
• Inference C (From 1 & Inference A): Since All $SP$ are $ES$, and Most $SP$ are $\neg PB$, we get: $ES \xleftrightarrow{\text{some}} \neg PB$ ("Some equity shareholders do not work on pro bono cases").