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Metamath Proof Explorer

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 ) )