Metamath Proof Explorer


Theorem axregszf

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

Ref Expression
Assertion axregszf ( 𝐴 ≠ ∅ → ∃ 𝑥 ∈ 𝐴 ( 𝑥 ∩ 𝐴 ) = ∅ )

Proof

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