Metamath Proof Explorer


Theorem axregszf

Description: Derivation of zfregs using ax-regs . (Contributed by BTernaryTau, 30-Dec-2025)

Ref Expression
Assertion axregszf ( 𝐴 ≠ ∅ → ∃ 𝑥𝐴 ( 𝑥𝐴 ) = ∅ )

Proof

Step Hyp Ref Expression
1 n0 ( 𝐴 ≠ ∅ ↔ ∃ 𝑥 𝑥𝐴 )
2 axregscl ( ∃ 𝑥 𝑥𝐴 → ∃ 𝑥 ( 𝑥𝐴 ∧ ∀ 𝑦 ( 𝑦𝑥 → ¬ 𝑦𝐴 ) ) )
3 disj1 ( ( 𝑥𝐴 ) = ∅ ↔ ∀ 𝑦 ( 𝑦𝑥 → ¬ 𝑦𝐴 ) )
4 3 rexbii ( ∃ 𝑥𝐴 ( 𝑥𝐴 ) = ∅ ↔ ∃ 𝑥𝐴𝑦 ( 𝑦𝑥 → ¬ 𝑦𝐴 ) )
5 df-rex ( ∃ 𝑥𝐴𝑦 ( 𝑦𝑥 → ¬ 𝑦𝐴 ) ↔ ∃ 𝑥 ( 𝑥𝐴 ∧ ∀ 𝑦 ( 𝑦𝑥 → ¬ 𝑦𝐴 ) ) )
6 4 5 bitr2i ( ∃ 𝑥 ( 𝑥𝐴 ∧ ∀ 𝑦 ( 𝑦𝑥 → ¬ 𝑦𝐴 ) ) ↔ ∃ 𝑥𝐴 ( 𝑥𝐴 ) = ∅ )
7 2 6 sylib ( ∃ 𝑥 𝑥𝐴 → ∃ 𝑥𝐴 ( 𝑥𝐴 ) = ∅ )
8 1 7 sylbi ( 𝐴 ≠ ∅ → ∃ 𝑥𝐴 ( 𝑥𝐴 ) = ∅ )