Theorem ac3 10415

Description: Axiom of Choice using abbreviations. The logical equivalence to ax-ac 10412 can be established by chaining aceq0 10071 and aceq2 10072. A standard textbook version of AC is derived from this one in dfac2a 10083, and this version of AC is derived from the textbook version in dfac2b 10084, showing their logical equivalence (see dfac2 10085).

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 9571. The key theorem for this (used in the proof of dfac2b 10084) is preleq 9569. 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 10414	. 2 ⊢ ∃𝑦∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑧 ∃!𝑣 ∈ 𝑧 ∃𝑢 ∈ 𝑦 (𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢)
2		aceq2 10072	. 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 1779   ≠ wne 2925  ∀wral 3044  ∃wrex 3053  ∃!wreu 3352  ∅c0 4296

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2701  ax-ac 10412

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2533  df-eu 2562  df-clab 2708  df-cleq 2721  df-clel 2803  df-ne 2926  df-ral 3045  df-rex 3054  df-rmo 3354  df-reu 3355  df-dif 3917  df-nul 4297

This theorem is referenced by:  axac2  10419