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Predicate Logic · step 11 of 13

Proving ∀-statements with an arbitrary element

To prove "for all x in D, P(x)" you take an arbitrary but fixed element of D, prove P of it using nothing specific to your choice, and then generalise. The word "arbitrary" is doing real work: the proof is valid precisely because no property of the chosen element beyond membership in D was ever used. This is the universal-introduction rule dressed in prose.

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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.

  1. Open with the fixing sentenceWrite "Let x ∈ D be arbitrary." This one line is worth a mark in essentially every scheme.
  2. Use only the defining propertiesEvery step must follow from x ∈ D and the hypotheses. If you divide by x, you have assumed x ≠ 0 and must justify it.
  3. Derive P(x) in fullComplete the argument for the arbitrary element, with each algebraic or logical step justified.
  4. Generalise explicitlyConclude "since x was arbitrary, ∀x ∈ D, P(x)." Skipping this line leaves the proof formally incomplete.

Worked example

Prove that for every integer n, n² + n is even.

  1. Let n ∈ ℤ be arbitrary.
  2. Factor: n² + n = n(n + 1).
  3. n and n + 1 are consecutive integers, so exactly one of them is even; write that one as 2k.
  4. Then n(n + 1) is 2k times an integer, hence of the form 2m with m ∈ ℤ.
  5. So n² + n is even. Since n was arbitrary, the claim holds for all integers.

Answer. n² + n = n(n + 1) is a product of consecutive integers, hence even, for every n ∈ ℤ.

Where marks get dropped

These are the specific errors that cost credit on proving ∀-statements with an arbitrary element questions — QED's rubric penalises each of them separately.

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Marked like an examiner

Every answer is scored against a point-by-point rubric with partial credit, so you see exactly which step of the method broke down — not just a tick or a cross.

Answer in real notation

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Proving ∀-statements with an arbitrary element — frequently asked questions

Why can’t I just check lots of cases?

Because a universal claim over an infinite domain has infinitely many cases. Checking a hundred integers leaves the other infinitely many untested — and there are famous claims that first fail at enormous numbers.

What if the domain is finite?

Then exhaustive checking IS a valid proof, and it is called proof by exhaustion. State that the domain is finite and check every element.

When should I use induction instead?

When the domain is ℕ and P(n + 1) is naturally derived from P(n). The arbitrary-element method suits claims that follow directly from definitions with no recursive structure.

The rest of Predicate Logic

Quantifiers, predicates, binding, and validity. Each subtopic below has its own method, worked example and mark-losing traps.

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