Theorem abbibw 43437

Description: Replace ax-10 2175, ax-11 2191, ax-12 2212 in abbib 2831 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 2755	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		vex 3458	. . . . 5 ⊢ 𝑦 ∈ V
3		abbibw.ph	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜃))
4	2, 3	elab 3637	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜃)
5		abbibw.ps	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))
6	2, 5	elab 3637	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒)
7	4, 6	bibi12i 342	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜃 ↔ 𝜒))
8	7	albii 1848	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑦(𝜃 ↔ 𝜒))
9	3, 5	bibi12d 348	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝜑 ↔ 𝜓) ↔ (𝜃 ↔ 𝜒)))
10	9	bicomd 226	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
11	10	equcoms 2049	. . 3 ⊢ (𝑦 = 𝑥 → ((𝜃 ↔ 𝜒) ↔ (𝜑 ↔ 𝜓)))
12	11	cbvalvw 2065	. 2 ⊢ (∀𝑦(𝜃 ↔ 𝜒) ↔ ∀𝑥(𝜑 ↔ 𝜓))
13	1, 8, 12	3bitri 300	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1567   = wceq 1569   ∈ wcel 2142  {cab 2740

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3456

This theorem is used by:  absnw  43438