Theorem rsp3eq 39300

Description: From a restricted universal statement over 𝐴, specialize to an arbitrary element class, cf. rsp3 39299. (Contributed by Peter Mazsa, 9-Feb-2026.)

Hypotheses

Ref	Expression
rsp3.1	⊢ Ⅎ𝑥𝐴
rsp3.2	⊢ Ⅎ𝑦𝐴
rsp3.3	⊢ Ⅎ𝑦𝜑
rsp3.4	⊢ Ⅎ𝑥𝜓
rsp3.5	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
rsp3eq	⊢ (∀𝑥 ∈ 𝐴 𝜑 → ((𝑦 = 𝐵 ∧ 𝐵 ∈ 𝐴) → 𝜓))

Distinct variable group:   𝑥,𝑦

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

Proof of Theorem rsp3eq

Step	Hyp	Ref	Expression
1		eqeltr 39173	. 2 ⊢ ((𝑦 = 𝐵 ∧ 𝐵 ∈ 𝐴) → 𝑦 ∈ 𝐴)
2		rsp3.1	. . 3 ⊢ Ⅎ𝑥𝐴
3		rsp3.2	. . 3 ⊢ Ⅎ𝑦𝐴
4		rsp3.3	. . 3 ⊢ Ⅎ𝑦𝜑
5		rsp3.4	. . 3 ⊢ Ⅎ𝑥𝜓
6		rsp3.5	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
7	2, 3, 4, 5, 6	rsp3 39299	. 2 ⊢ (∀𝑥 ∈ 𝐴 𝜑 → (𝑦 ∈ 𝐴 → 𝜓))
8	1, 7	syl5 35	1 ⊢ (∀𝑥 ∈ 𝐴 𝜑 → ((𝑦 = 𝐵 ∧ 𝐵 ∈ 𝐴) → 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569  Ⅎwnf 1815   ∈ wcel 2144  Ⅎwnfc 2907  ∀wral 3076

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-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ral 3077

This theorem is used by: (None)