Natural deduction with quantifier rules
Predicate natural deduction adds four rules to the propositional ones. ∀E instantiates a universal at any term; ∃I generalises a witness to an existential. The two harder rules carry side conditions: ∀I requires the variable to be genuinely arbitrary, and ∃E requires you to reason from a fresh name inside a box. Nearly all lost marks here come from violating those conditions rather than from the algebra.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- ∀E — instantiate freelyFrom ∀x φ(x) infer φ(t) for any term t. There is no restriction, which makes this the safest rule.
- ∀I — check arbitrarinessFrom φ(a) infer ∀x φ(x) only if a is fresh: it must not appear in any premise or undischarged assumption. State that it is arbitrary.
- ∃I — supply a witnessFrom φ(t) infer ∃x φ(x). Name the term you used; that is the mark.
- ∃E — reason under a fresh nameFrom ∃x φ(x), assume φ(a) for a fresh a, derive a conclusion C that does not mention a, then discharge. Both freshness and the absence of a in C are required.
Worked example
Prove ∃x (P(x) ∧ Q(x)) ⊢ ∃x P(x).
- 1. ∃x (P(x) ∧ Q(x)) (premise).
- 2. Assume P(a) ∧ Q(a) for a fresh name a — open the ∃E box.
- 3. P(a), by ∧E on line 2.
- 4. ∃x P(x), by ∃I on line 3 with witness a.
- 5. Close the box: ∃x P(x), by ∃E on lines 1 and 2–4. The conclusion does not mention a, so the rule applies.
Answer. ∃x P(x) is derived in five lines, with a used only inside the ∃E box.
Where marks get dropped
These are the specific errors that cost credit on natural deduction with quantifier rules questions — QED's rubric penalises each of them separately.
- Applying ∀I to a name that occurs in an undischarged assumption. That "proves" universal claims from single instances and invalidates the whole derivation.
- Letting the fresh name escape an ∃E box into the conclusion. If C mentions a, the rule does not apply.
- Using ∃E as though it were ∃-elimination to a specific object. You never learn which object it was — only that reasoning works for an arbitrary one.
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Natural deduction with quantifier rules — frequently asked questions
Why does ∀I need a freshness condition?
Without it you could assume P(a), infer ∀x P(x), and prove that one instance implies a universal law. The condition encodes "a was arbitrary" — the exact step a mathematician makes informally.
Can I prove ∀x ∃y R(x,y) from ∃y ∀x R(x,y)?
Yes: use ∃E to name the universal witness, ∀E to instantiate at an arbitrary a, ∃I to reintroduce, then ∀I. The reverse direction cannot be derived, and no correct rule application gets you there.
How are these proofs marked?
Per line: the formula, the rule name, the line references, and the side condition where one applies. QED marks the side conditions as their own criteria because that is where real errors hide.
The rest of Predicate Logic
Quantifiers, predicates, binding, and validity. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Universal & existential quantifiers
- 2Translating English with predicates
- 3Free vs bound variables & scope
- 4Negating quantified statements
- 5Validity & counter-models
- 6Nested quantifiers & quantifier order
- 7Prenex normal form
- 8Interpretations, structures & satisfaction
- 9Equality & uniqueness (∃!)
- 10Natural deduction with quantifier rules
- 11Proving ∀-statements with an arbitrary element
- 12Disproving with a single counterexample
- 13Skolemisation & clausal form
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