Theorem axac2

Description: Derive ax-ac2 from ax-ac . (Contributed by NM, 19-Dec-2016) (New usage is discouraged.) (Proof modification is discouraged.)

	Ref	Expression
Assertion	axac2	⊢ ∃ 𝑦 ∀ 𝑧 ∃ 𝑣 ∀ 𝑢 ( ( 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑦 → ( ( 𝑣 ∈ 𝑥 ∧ ¬ 𝑦 = 𝑣 ) ∧ 𝑧 ∈ 𝑣 ) ) ) ∨ ( ¬ 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑥 → ( ( 𝑣 ∈ 𝑧 ∧ 𝑣 ∈ 𝑦 ) ∧ ( ( 𝑢 ∈ 𝑧 ∧ 𝑢 ∈ 𝑦 ) → 𝑢 = 𝑣 ) ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		dfac2a	⊢ ( ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∈ 𝑥 ( 𝑧 ≠ ∅ → ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) ) → CHOICE )
2		ac3	⊢ ∃ 𝑦 ∀ 𝑧 ∈ 𝑥 ( 𝑧 ≠ ∅ → ∃! 𝑣 ∈ 𝑧 ∃ 𝑢 ∈ 𝑦 ( 𝑧 ∈ 𝑢 ∧ 𝑣 ∈ 𝑢 ) )
3	1 2	mpg	⊢ CHOICE
4		dfackm	⊢ ( CHOICE ↔ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∃ 𝑣 ∀ 𝑢 ( ( 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑦 → ( ( 𝑣 ∈ 𝑥 ∧ ¬ 𝑦 = 𝑣 ) ∧ 𝑧 ∈ 𝑣 ) ) ) ∨ ( ¬ 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑥 → ( ( 𝑣 ∈ 𝑧 ∧ 𝑣 ∈ 𝑦 ) ∧ ( ( 𝑢 ∈ 𝑧 ∧ 𝑢 ∈ 𝑦 ) → 𝑢 = 𝑣 ) ) ) ) ) )
5	3 4	mpbi	⊢ ∀ 𝑥 ∃ 𝑦 ∀ 𝑧 ∃ 𝑣 ∀ 𝑢 ( ( 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑦 → ( ( 𝑣 ∈ 𝑥 ∧ ¬ 𝑦 = 𝑣 ) ∧ 𝑧 ∈ 𝑣 ) ) ) ∨ ( ¬ 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑥 → ( ( 𝑣 ∈ 𝑧 ∧ 𝑣 ∈ 𝑦 ) ∧ ( ( 𝑢 ∈ 𝑧 ∧ 𝑢 ∈ 𝑦 ) → 𝑢 = 𝑣 ) ) ) ) )
6	5	spi	⊢ ∃ 𝑦 ∀ 𝑧 ∃ 𝑣 ∀ 𝑢 ( ( 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑦 → ( ( 𝑣 ∈ 𝑥 ∧ ¬ 𝑦 = 𝑣 ) ∧ 𝑧 ∈ 𝑣 ) ) ) ∨ ( ¬ 𝑦 ∈ 𝑥 ∧ ( 𝑧 ∈ 𝑥 → ( ( 𝑣 ∈ 𝑧 ∧ 𝑣 ∈ 𝑦 ) ∧ ( ( 𝑢 ∈ 𝑧 ∧ 𝑢 ∈ 𝑦 ) → 𝑢 = 𝑣 ) ) ) ) )