Theorem abbibw 42632

Description: Replace ax-10 2141, ax-11 2158, ax-12 2178 in abbib 2814 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 2733	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		vex 3492	. . . . 5 ⊢ 𝑦 ∈ V
3		abbibw.ph	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜃))
4	2, 3	elab 3694	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜃)
5		abbibw.ps	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
6	2, 5	elab 3694	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒)
7	4, 6	bibi12i 339	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜃 ↔ 𝜒))
8	7	albii 1817	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑦(𝜃 ↔ 𝜒))
9	3, 5	bibi12d 345	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝜑 ↔ 𝜓) ↔ (𝜃 ↔ 𝜒)))
10	9	bicomd 223	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
11	10	equcoms 2019	. . 3 ⊢ (𝑦 = 𝑥 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
12	11	cbvalvw 2035	. 2 ⊢ (∀𝑦(𝜃 ↔ 𝜒) ↔ ∀𝑥(𝜑 ↔ 𝜓))
13	1, 8, 12	3bitri 297	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

Syntax hints:   ↔ wb 206  ∀wal 1535   = wceq 1537   ∈ wcel 2108  {cab 2717

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-v 3490

This theorem is referenced by:  absnw  42633