Theorem bj-gabeqis 37773

Description: Equality of generalized class abstractions, with implicit substitution. (Contributed by BJ, 4-Oct-2024.)

Hypotheses

Ref	Expression
bj-gabeqis.c	⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
bj-gabeqis.f	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
bj-gabeqis	⊢ {𝐴 ∣ 𝑥 ∣ 𝜑} = {𝐵 ∣ 𝑦 ∣ 𝜓}

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

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

Proof of Theorem bj-gabeqis

Dummy variables 𝑢 𝑣 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		bj-gabeqis.c	. . . . . . 7 ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)
2	1	adantl 487	. . . . . 6 ⊢ ((𝑢 = 𝑣 ∧ 𝑥 = 𝑦) → 𝐴 = 𝐵)
3		simpl 488	. . . . . 6 ⊢ ((𝑢 = 𝑣 ∧ 𝑥 = 𝑦) → 𝑢 = 𝑣)
4	2, 3	eqeq12d 2776	. . . . 5 ⊢ ((𝑢 = 𝑣 ∧ 𝑥 = 𝑦) → (𝐴 = 𝑢 ↔ 𝐵 = 𝑣))
5		bj-gabeqis.f	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
6	5	adantl 487	. . . . 5 ⊢ ((𝑢 = 𝑣 ∧ 𝑥 = 𝑦) → (𝜑 ↔ 𝜓))
7	4, 6	anbi12d 644	. . . 4 ⊢ ((𝑢 = 𝑣 ∧ 𝑥 = 𝑦) → ((𝐴 = 𝑢 ∧ 𝜑) ↔ (𝐵 = 𝑣 ∧ 𝜓)))
8	7	cbvexdvaw 2072	. . 3 ⊢ (𝑢 = 𝑣 → (∃𝑥(𝐴 = 𝑢 ∧ 𝜑) ↔ ∃𝑦(𝐵 = 𝑣 ∧ 𝜓)))
9	8	cbvabv 2830	. 2 ⊢ {𝑢 ∣ ∃𝑥(𝐴 = 𝑢 ∧ 𝜑)} = {𝑣 ∣ ∃𝑦(𝐵 = 𝑣 ∧ 𝜓)}
10		df-bj-gab 37769	. 2 ⊢ {𝐴 ∣ 𝑥 ∣ 𝜑} = {𝑢 ∣ ∃𝑥(𝐴 = 𝑢 ∧ 𝜑)}
11		df-bj-gab 37769	. 2 ⊢ {𝐵 ∣ 𝑦 ∣ 𝜓} = {𝑣 ∣ ∃𝑦(𝐵 = 𝑣 ∧ 𝜓)}
12	9, 10, 11	3eqtr4i 2793	1 ⊢ {𝐴 ∣ 𝑥 ∣ 𝜑} = {𝐵 ∣ 𝑦 ∣ 𝜓}

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812  {cab 2738  {bj-cgab 37768

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-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-bj-gab 37769

This theorem is used by: (None)