Theorem abbibw 42237

Description: Replace ax-10 2129, ax-11 2146, ax-12 2166 in abbib 2797 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 2718	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		vex 3465	. . . . 5 ⊢ 𝑦 ∈ V
3		abbibw.ph	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜃))
4	2, 3	elab 3664	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜃)
5		abbibw.ps	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
6	2, 5	elab 3664	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒)
7	4, 6	bibi12i 338	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜃 ↔ 𝜒))
8	7	albii 1813	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑦(𝜃 ↔ 𝜒))
9	3, 5	bibi12d 344	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝜑 ↔ 𝜓) ↔ (𝜃 ↔ 𝜒)))
10	9	bicomd 222	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
11	10	equcoms 2015	. . 3 ⊢ (𝑦 = 𝑥 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
12	11	cbvalvw 2031	. 2 ⊢ (∀𝑦(𝜃 ↔ 𝜒) ↔ ∀𝑥(𝜑 ↔ 𝜓))
13	1, 8, 12	3bitri 296	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

Syntax hints:   ↔ wb 205  ∀wal 1531   = wceq 1533   ∈ wcel 2098  {cab 2702

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2696

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2703  df-cleq 2717  df-clel 2802  df-v 3463

This theorem is referenced by:  absnw  42238