CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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.

If ZF is consistent, DMC is not provable in ZF

Statement

If $Z F$ is consistent, then $Z F$ does not prove DMC (Dependent multiple choice in finite-level tree form); indeed there is a model of $Z F$ with countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) in which DMC fails.

Facts & Assumptions

Given: The assumed consistency of $Z F$ .

[F1]

Relative to $Con (Z F)$ , the theory $Z F + {A C}_{ω} + \neg U R Y$ is consistent (Relative consistency of Countable Choice without Urysohn's lemma).

[F2]

DMC implies Urysohn's lemma over $Z F$ (DMC implies Urysohn's lemma).

Proof

technique · contradiction

1.1

Assume $Con (Z F)$ and suppose $Z F$ proves DMC.

assume-contragiven

2.1

Then $Z F + {A C}_{ω}$ proves DMC, hence by [F2] proves $U R Y$ ; but by [F1] the theory $Z F + {A C}_{ω} + \neg U R Y$ is consistent, and it would prove both $U R Y$ and its negation, hence be inconsistent.

step 1.1F1F2

3.1

This contradiction shows that $Z F$ does not prove DMC, conditionally on $Con (Z F)$ ; the witness model supplied by [F1] has countable choice, while Urysohn's lemma fails there and therefore DMC fails.

step 2.1F1F2discharge-contradiction∎

Depends on

Used by

Dependency tree · two levels

25 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)