Theorem ac3 10464

Description: Axiom of Choice using abbreviations. The logical equivalence to ax-ac 10461 can be established by chaining aceq0 10121 and aceq2 10122. A standard textbook version of AC is derived from this one in dfac2a 10132, and this version of AC is derived from the textbook version in dfac2b 10133, showing their logical equivalence (see dfac2 10134).

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 9597. The key theorem for this (used in the proof of dfac2b 10133) is preleq 9595. 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 10463	. 2 ⊢ ∃𝑦∀𝑧 ∈ 𝑥 ∀𝑤 ∈ 𝑧 ∃!𝑣 ∈ 𝑧 ∃𝑢 ∈ 𝑦 (𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢)
2		aceq2 10122	. 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 2955  ∀wral 3076  ∃wrex 3086  ∃!wreu 3363  ∅c0 4279

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 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732  ax-ac 10461

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 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rex 3087  df-rmo 3365  df-reu 3366  df-dif 3902  df-nul 4280

This theorem is used by:  axac2  10468