Theorem abbibw 43260

Description: Replace ax-10 2176, ax-11 2192, ax-12 2213 in abbib 2832 with substitution hypotheses. (Contributed by SN, 27-May-2025.)

Hypotheses

Ref	Expression
abbibw.ph	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜃))
abbibw.ps	⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
abbibw	⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

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

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)   𝜒(𝑦)   𝜃(𝑦)

Proof of Theorem abbibw

Step	Hyp	Ref	Expression
1		dfcleq 2756	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		vex 3459	. . . . 5 ⊢ 𝑦 ∈ V
3		abbibw.ph	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜃))
4	2, 3	elab 3639	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜃)
5		abbibw.ps	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
6	2, 5	elab 3639	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒)
7	4, 6	bibi12i 341	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜃 ↔ 𝜒))
8	7	albii 1840	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑦(𝜃 ↔ 𝜒))
9	3, 5	bibi12d 347	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝜑 ↔ 𝜓) ↔ (𝜃 ↔ 𝜒)))
10	9	bicomd 225	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
11	10	equcoms 2041	. . 3 ⊢ (𝑦 = 𝑥 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
12	11	cbvalvw 2057	. 2 ⊢ (∀𝑦(𝜃 ↔ 𝜒) ↔ ∀𝑥(𝜑 ↔ 𝜓))
13	1, 8, 12	3bitri 299	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

Syntax hints:   ↔ wb 208  ∀wal 1559   = wceq 1561   ∈ wcel 2143  {cab 2741

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1816  ax-4 1830  ax-5 1931  ax-6 1988  ax-7 2029  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1564  df-ex 1801  df-sb 2092  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3457

This theorem is referenced by:  absnw  43261