CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A failure of Urysohn separation forces failure of DMC

Statement

Over ZF, failure of Urysohn's lemma implies failure of DMC (Dependent multiple choice in finite-level tree form). In particular, if any consistent theory $T$ extends ZF and includes $\neg U R Y$ , then ZF does not prove DMC. No existence or consistency of such a $T$ is asserted.

Facts & Assumptions

Given: ZF, with $\neg U R Y$ for the first claim; and a consistent extending theory $T$ containing $\neg U R Y$ for the second claim.

[F1]

Over ZF, DMC implies Urysohn's lemma (DMC implies Urysohn's lemma).

Proof

1.1givenF1

Under $\neg U R Y$ , DMC would give $U R Y$ by F1, a contradiction. Therefore $\neg D M C$ holds. This proves the internal implication over ZF.

2.1givenF1step 1.1∎

If ZF proved DMC, the same finite derivation would hold in the extending $T$ , and the finite ZF proof F1 would give $U R Y$ in $T$ . Together with its axiom $\neg U R Y$ this contradicts consistency of $T$ . Thus ZF cannot prove DMC under the stated additional consistency hypothesis.

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Eleftherios Tachtsis, The Urysohn Lemma is independent of ZF + Countable Choice (standard reference, not scraped)
Eleftherios Tachtsis, Erratum to The Urysohn Lemma is independent of ZF + Countable Choice (standard reference, not scraped)