Theorem axac

Description: Derive ax-ac from ax-ac2 . Note that ax-reg is used by the proof. (Contributed by NM, 19-Dec-2016) (Proof modification is discouraged.)

	Ref	Expression
Assertion	axac	⊢ ∃ 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		axac3	⊢ CHOICE
2		dfac0	⊢ ( CHOICE ↔ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) → ∃ 𝑣 ∀ 𝑢 ( ∃ 𝑡 ( ( 𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡 ) ∧ ( 𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦 ) ) ↔ 𝑢 = 𝑣 ) ) )
3	1 2	mpbi	⊢ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) → ∃ 𝑣 ∀ 𝑢 ( ∃ 𝑡 ( ( 𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡 ) ∧ ( 𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦 ) ) ↔ 𝑢 = 𝑣 ) )
4	3	spi	⊢ ∃ 𝑦 ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 ∈ 𝑤 ∧ 𝑤 ∈ 𝑥 ) → ∃ 𝑣 ∀ 𝑢 ( ∃ 𝑡 ( ( 𝑢 ∈ 𝑤 ∧ 𝑤 ∈ 𝑡 ) ∧ ( 𝑢 ∈ 𝑡 ∧ 𝑡 ∈ 𝑦 ) ) ↔ 𝑢 = 𝑣 ) )