Theorem sepgi 5251

Description: Inference associated with sepg 5250. The requirement that 𝑦 not occur in 𝜑 is necessary, as notsep 5324 shows. (Contributed by NM, 21-Jun-1993.) (Revised by BJ, 14-Jul-2026.)

Hypothesis

Ref	Expression
sepgi.1	⊢ 𝐴 ∈ V

Assertion

Ref	Expression
sepgi	⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))

Distinct variable groups:   𝑥,𝐴,𝑦   𝜑,𝑦

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem sepgi

Step	Hyp	Ref	Expression
1		sepgi.1	. 2 ⊢ 𝐴 ∈ V
2		sepg 5250	. 2 ⊢ (𝐴 ∈ V → ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
3	1, 2	ax-mp 5	1 ⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401  ∀wal 1567  ∃wex 1811   ∈ wcel 2144  Vcvv 3450

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835

This theorem is used by:  inex1  5276