Theorem ac3 10500

Description: Axiom of Choice using abbreviations. The logical equivalence to ax-ac 10497 can be established by chaining aceq0 10156 and aceq2 10157. A standard textbook version of AC is derived from this one in dfac2a 10168, and this version of AC is derived from the textbook version in dfac2b 10169, showing their logical equivalence (see dfac2 10170).

The following sketch will help you understand this version of the axiom. Given any set 𝑥, the axiom says that there exists a 𝑦 that is a collection of unordered pairs, one pair for each nonempty member of 𝑥. One entry in the pair is the member of 𝑥, and the other entry is some arbitrary member of that member of 𝑥. Using the Axiom of Regularity, we can show that 𝑦 is really a set of ordered pairs, very similar to the ordered pair construction opthreg 9656. The key theorem for this (used in the proof of dfac2b 10169) is preleq 9654. With this modified definition of ordered pair, it can be seen that 𝑦 is actually a choice function on the members of 𝑥.

For example, suppose 𝑥 = {{1, 2}, {1, 3}, {2, 3, 4}}. Let us try 𝑦 = {{{1, 2}, 1}, {{1, 3}, 1}, {{2, 3, 4}, 2}}. For the member (of 𝑥) 𝑧 = {1, 2}, the only assignment to 𝑤 and 𝑣 that satisfies the axiom is 𝑤 = 1 and 𝑣 = {{1, 2}, 1}, so there is exactly one 𝑤 as required. We verify the other two members of 𝑥 similarly. Thus, 𝑦 satisfies the axiom. Using our modified ordered pair definition, we can say that 𝑦 corresponds to the choice function {⟨{1, 2}, 1⟩, ⟨{1, 3}, 1⟩, ⟨{2, 3, 4}, 2⟩}. Of course other choices for 𝑦 will also satisfy the axiom, for example 𝑦 = {{{1, 2}, 2}, {{1, 3}, 1}, {{2, 3, 4}, 4}}. What AC tells us is that there exists at least one such 𝑦, but it doesn't tell us which one.

(New usage is discouraged.) (Contributed by NM, 19-Jul-1996.)

Assertion

Ref	Expression
ac3	⊢ ∃𝑦∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → ∃!𝑤 ∈ 𝑧 ∃𝑣 ∈ 𝑦 (𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣))

Distinct variable group:   𝑥,𝑦,𝑧,𝑤,𝑣

Proof of Theorem ac3

Dummy variable 𝑢 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ac2 10499	. 2 ⊢ ∃𝑦∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑧 ∃!𝑣 ∈ 𝑧 ∃𝑢 ∈ 𝑦 (𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢)
2		aceq2 10157	. 2 ⊢ (∃𝑦∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑧 ∃!𝑣 ∈ 𝑧 ∃𝑢 ∈ 𝑦 (𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢) ↔ ∃𝑦∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → ∃!𝑤 ∈ 𝑧 ∃𝑣 ∈ 𝑦 (𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣)))
3	1, 2	mpbi 230	1 ⊢ ∃𝑦∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → ∃!𝑤 ∈ 𝑧 ∃𝑣 ∈ 𝑦 (𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣))

Syntax hints:   → wi 4   ∧ wa 395  ∃wex 1776   ≠ wne 2938  ∀wral 3059  ∃wrex 3068  ∃!wreu 3376  ∅c0 4339

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706  ax-ac 10497

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-fal 1550  df-ex 1777  df-nf 1781  df-sb 2063  df-mo 2538  df-eu 2567  df-clab 2713  df-cleq 2727  df-clel 2814  df-ne 2939  df-ral 3060  df-rex 3069  df-rmo 3378  df-reu 3379  df-dif 3966  df-nul 4340

This theorem is referenced by:  axac2  10504