Theorem bj-dfsbc 37473

Description: Proof of df-sbc 3739 when taking bj-df-sb 37471 as definition. (Contributed by BJ, 19-Feb-2026.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
bj-dfsbc	⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ [𝐴 / 𝑥]𝜑)

Proof of Theorem bj-dfsbc

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-clab 2739	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
2		sb6 2122	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3	1, 2	bitri 278	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
4	3	anbi2i 635	. . 3 ⊢ ((𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ (𝑦 = 𝐴 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
5	4	exbii 1881	. 2 ⊢ (∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ ∃𝑦(𝑦 = 𝐴 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
6		dfclel 2836	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}))
7		bj-df-sb 37471	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑦(𝑦 = 𝐴 ∧ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
8	5, 6, 7	3bitr4i 306	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ [𝐴 / 𝑥]𝜑)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570  ∃wex 1812  [wsb 2099   ∈ wcel 2145  {cab 2738  [wsbc 3738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-12 2213  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-sbc 3739

This theorem is used by: (None)