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Metamath Proof Explorer

Theorem aceq2

Description: Equivalence of two versions of the Axiom of Choice. The proof uses neither AC nor the Axiom of Regularity. (Contributed by NM, 5-Apr-2004)

Ref Expression
Assertion aceq2 ( ∃ 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 ) ) )

Proof

Step Hyp Ref Expression
1 df-ral ⊢ ( ∀ 𝑡 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ∀ 𝑡 ( 𝑡 ∈ 𝑧 → ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) )
2 19.23v ⊢ ( ∀ 𝑡 ( 𝑡 ∈ 𝑧 → ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) ↔ ( ∃ 𝑡 𝑡 ∈ 𝑧 → ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) )
3 1 2 bitri ⊢ ( ∀ 𝑡 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ( ∃ 𝑡 𝑡 ∈ 𝑧 → ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) )
4 biidd ⊢ ( 𝑤 = 𝑡 → ( ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) )
5 4 cbvralvw ⊢ ( ∀ 𝑤 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ∀ 𝑡 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) )
6 n0 ⊢ ( 𝑧 ≠ ∅ ↔ ∃ 𝑡 𝑡 ∈ 𝑧 )
7 elequ2 ⊢ ( 𝑣 = 𝑢 → ( 𝑧 ∈ 𝑣 ↔ 𝑧 ∈ 𝑢 ) )
8 elequ2 ⊢ ( 𝑣 = 𝑢 → ( 𝑤 ∈ 𝑣 ↔ 𝑤 ∈ 𝑢 ) )
9 7 8 anbi12d ⊢ ( 𝑣 = 𝑢 → ( ( 𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣 ) ↔ ( 𝑧 ∈ 𝑢 ∧ 𝑤 ∈ 𝑢 ) ) )
10 9 cbvrexvw ⊢ ( ∃ 𝑣 ∈ 𝑦 ( 𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣 ) ↔ ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑤 ∈ 𝑢 ) )
11 10 reubii ⊢ ( ∃! 𝑤 ∈ 𝑧 ∃ 𝑣 ∈ 𝑦 ( 𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣 ) ↔ ∃! 𝑤 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑤 ∈ 𝑢 ) )
12 elequ1 ⊢ ( 𝑤 = 𝑣 → ( 𝑤 ∈ 𝑢 ↔ 𝑣 ∈ 𝑢 ) )
13 12 anbi2d ⊢ ( 𝑤 = 𝑣 → ( ( 𝑧 ∈ 𝑢 ∧ 𝑤 ∈ 𝑢 ) ↔ ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) )
14 13 rexbidv ⊢ ( 𝑤 = 𝑣 → ( ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑤 ∈ 𝑢 ) ↔ ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) )
15 14 cbvreuvw ⊢ ( ∃! 𝑤 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑤 ∈ 𝑢 ) ↔ ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) )
16 11 15 bitri ⊢ ( ∃! 𝑤 ∈ 𝑧 ∃ 𝑣 ∈ 𝑦 ( 𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣 ) ↔ ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) )
17 6 16 imbi12i ⊢ ( ( 𝑧 ≠ ∅ → ∃! 𝑤 ∈ 𝑧 ∃ 𝑣 ∈ 𝑦 ( 𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣 ) ) ↔ ( ∃ 𝑡 𝑡 ∈ 𝑧 → ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) )
18 3 5 17 3bitr4i ⊢ ( ∀ 𝑤 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ( 𝑧 ≠ ∅ → ∃! 𝑤 ∈ 𝑧 ∃ 𝑣 ∈ 𝑦 ( 𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣 ) ) )
19 18 ralbii ⊢ ( ∀ 𝑧 ∈ 𝑥 ∀ 𝑤 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ∀ 𝑧 ∈ 𝑥 ( 𝑧 ≠ ∅ → ∃! 𝑤 ∈ 𝑧 ∃ 𝑣 ∈ 𝑦 ( 𝑧 ∈ 𝑣 ∧ 𝑤 ∈ 𝑣 ) ) )
20 19 exbii ⊢ ( ∃ 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 ) ) )