Symbolic LogicPredicate Logic: Proofs

Universal Instantiation and Existential Generalization (UI, EG)

UI: from ∀x Fx, infer Fa for any name. EG: from Fa, infer ∃x Fx.

Not started Jump to practice

Worlds can show that a predicate argument is invalid (Counterexample Worlds), but to show validity we need proofs. The plan is simple: take the quantifiers off, reason with the eighteen rules you already know (Proofs with All Eighteen Rules), and put quantifiers back on at the end. This lesson gives the two easiest quantifier rules, universal instantiation and existential generalization, and uses Quantifier Negation when a premise starts with a negation.

The letters stand for any formulas: from $\forall x(Fx \to Gx)$, UI gives $Fa \to Ga$.
RuleFromInfer
UI (universal instantiation)$\forall x Fx$$Fa$ (any name), or $Fy$ (a variable)
EG (existential generalization)$Fa$$\exists x Fx$

Key idea

UI: from a universal statement, drop the $\forall x$ and put one name (or variable) for every $x$ it covered. EG: from a statement about a name, replace that name (in some or all of its places) with a variable and put $\exists$ in front of the whole line. Both apply only to whole lines.

Why it works

If something holds for everything, it holds for Ava, for Ben, for any particular thing you choose: that is UI. If something holds for Ava, then it holds for at least one thing: that is EG. Both are plainly truth-preserving. With them, "All dogs are mammals; Rex is a dog; so Rex is a mammal" becomes a three-line proof.
All dogs are mammals; Rex is a dog; so Rex is a mammal.

Worked examples

Worked example: UI with a compound inside

From $\forall x(Fx \to \neg Gx)$, what does UI give with the name $b$?
  1. Drop the quantifier.
  2. Put $b$ for every $x$.

Answer: $Fb \to \neg Gb$

Worked example: EG

From $Fa \land Ga$, what can EG give?
  1. Replace $a$ by $x$ and put $\exists x$ in front.
  2. EG may replace all the $a$'s or only some.

Answer: $\exists x(Fx \land Gx)$ (also allowed: $\exists x(Fx \land Ga)$).

Worked example: A short proof

Premises $\forall x(Fx \to Gx)$ and $Fc$. Prove $\exists x Gx$.
  1. Line 3: $Fc \to Gc$ by UI 1.
  2. Line 4: $Gc$ by MP 3, 2.
  3. Line 5: $\exists x Gx$ by EG 4.

Answer: UI, MP, EG.

Common mistake

Applying UI to part of a line, such as $\forall x Fx \to Ga$. Its main connective is $\to$, not $\forall$: the quantifier covers only $Fx$. Like the rules of inference, the quantifier rules act only on whole lines whose main connective is the quantifier. If a line begins with $\neg Ax$, use QN first to get a quantifier in front.

Choosing the name for UI is up to you, so choose with the goal in mind. If the conclusion or another premise mentions $c$, instantiate with $c$, because that is the object the argument is about. Instantiating a universal premise with a name nobody mentions is legal but rarely gets you anywhere.

  • UI: universal in front → any name or variable, the same one everywhere.
  • EG: a name → a variable under $\exists$, covering the whole line.
  • Never: UI on a part of a line, or EG giving $\forall$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

Loading practice…

Builds on

Leads to

Sources and further reading

  • Patrick J. Hurley and Lori Watson, A Concise Introduction to Logic, 13th ed., section 8.2, Using the Rules of Inference (UI, EG)
  • forall x: Calgary, an open introduction to formal logic (P. D. Magnus, Tim Button and others; Open Logic Project), chapter on basic rules for FOL
@teddyrookbook GitHub