Metamath Proof Explorer


Theorem dfac0

Description: Equivalence of two versions of the Axiom of Choice. The proof uses the Axiom of Regularity. The right-hand side is our original ax-ac . (Contributed by Mario Carneiro, 17-May-2015)

Ref Expression
Assertion dfac0 ( CHOICE ↔ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) → ∃ 𝑣 ∀ 𝑢 ( ∃ 𝑡 ( ( 𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡 ) ∧ ( 𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )

Proof

Step Hyp Ref Expression
1 dfac7 ⊢ ( CHOICE ↔ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∈ 𝑥 ∀ 𝑤 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) )
2 aceq0 ⊢ ( ∃ 𝑦 ∀ 𝑧 ∈ 𝑥 ∀ 𝑤 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ∃ 𝑦 ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) → ∃ 𝑣 ∀ 𝑢 ( ∃ 𝑡 ( ( 𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡 ) ∧ ( 𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )
3 2 albii ⊢ ( ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∈ 𝑥 ∀ 𝑤 ∈ 𝑧 ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ↔ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) → ∃ 𝑣 ∀ 𝑢 ( ∃ 𝑡 ( ( 𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡 ) ∧ ( 𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )
4 1 3 bitri ⊢ ( CHOICE ↔ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) → ∃ 𝑣 ∀ 𝑢 ( ∃ 𝑡 ( ( 𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡 ) ∧ ( 𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )