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 ↔ ∀ 𝑥𝑦𝑧𝑤 ( ( 𝑧𝑤𝑤𝑥 ) → ∃ 𝑣𝑢 ( ∃ 𝑡 ( ( 𝑢𝑤𝑤𝑡 ) ∧ ( 𝑢𝑡𝑡𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )

Proof

Step Hyp Ref Expression
1 dfac7 ( CHOICE ↔ ∀ 𝑥𝑦𝑧𝑥𝑤𝑧 ∃! 𝑣𝑧𝑢𝑦 ( 𝑧𝑢𝑣𝑢 ) )
2 aceq0 ( ∃ 𝑦𝑧𝑥𝑤𝑧 ∃! 𝑣𝑧𝑢𝑦 ( 𝑧𝑢𝑣𝑢 ) ↔ ∃ 𝑦𝑧𝑤 ( ( 𝑧𝑤𝑤𝑥 ) → ∃ 𝑣𝑢 ( ∃ 𝑡 ( ( 𝑢𝑤𝑤𝑡 ) ∧ ( 𝑢𝑡𝑡𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )
3 2 albii ( ∀ 𝑥𝑦𝑧𝑥𝑤𝑧 ∃! 𝑣𝑧𝑢𝑦 ( 𝑧𝑢𝑣𝑢 ) ↔ ∀ 𝑥𝑦𝑧𝑤 ( ( 𝑧𝑤𝑤𝑥 ) → ∃ 𝑣𝑢 ( ∃ 𝑡 ( ( 𝑢𝑤𝑤𝑡 ) ∧ ( 𝑢𝑡𝑡𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )
4 1 3 bitri ( CHOICE ↔ ∀ 𝑥𝑦𝑧𝑤 ( ( 𝑧𝑤𝑤𝑥 ) → ∃ 𝑣𝑢 ( ∃ 𝑡 ( ( 𝑢𝑤𝑤𝑡 ) ∧ ( 𝑢𝑡𝑡𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )