Theorem ac3 10457

Description: Axiom of Choice using abbreviations. The logical equivalence to ax-ac 10454 can be established by chaining aceq0 10114 and aceq2 10115. A standard textbook version of AC is derived from this one in dfac2a 10125, and this version of AC is derived from the textbook version in dfac2b 10126, showing their logical equivalence (see dfac2 10127).

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 9590. The key theorem for this (used in the proof of dfac2b 10126) is preleq 9588. 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 10456	. 2 ⊢ ∃𝑦∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑧 ∃!𝑣 ∈ 𝑧 ∃𝑢 ∈ 𝑦 (𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢)
2		aceq2 10115	. 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 233	1 ⊢ ∃𝑦∀𝑧 ∈ 𝑥 (𝑧 ≠ ∅ → ∃!𝑤 ∈ 𝑧 ∃𝑣 ∈ 𝑦 (𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∃wex 1812   ≠ wne 2960  ∀wral 3081  ∃wrex 3091  ∃!wreu 3369  ∅c0 4286

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737  ax-ac 10454

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2569  df-eu 2599  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-rex 3092  df-rmo 3371  df-reu 3372  df-dif 3909  df-nul 4287

This theorem is used by:  axac2  10461