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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 ) ) )