Theorem elabgw 3633

Description: Membership in a class abstraction, using two substitution hypotheses to avoid a disjoint variable condition on 𝑥 and 𝐴. This is to elabg 3632 what sbievw2 2136 is to sbievw 2132. (Contributed by SN, 20-Apr-2024.)

Hypotheses

Ref	Expression
elabgw.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
elabgw.2	⊢ (𝑦 = 𝐴 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
elabgw	⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜒))

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

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)   𝜒(𝑥)   𝐴(𝑥)   𝑉(𝑥, 𝑦)

Proof of Theorem elabgw

Step	Hyp	Ref	Expression
1		eleq1 2851	. 2 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
2		elabgw.2	. 2 ⊢ (𝑦 = 𝐴 → (𝜓 ↔ 𝜒))
3		df-clab 2742	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		elabgw.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	4	sbievw 2132	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
6	3, 5	bitri 278	. 2 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
7	1, 2, 6	vtoclbg 3522	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  [wsb 2100   ∈ wcel 2146  {cab 2741

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838

This theorem is used by:  elab2gw  3634