Theorem abbib 2799

Description: Equal class abstractions require equivalent formulas, and conversely. (Contributed by NM, 25-Nov-2013.) (Revised by Mario Carneiro, 11-Aug-2016.) Remove dependency on ax-8 2111 and df-clel 2804 (by avoiding use of cleqh 2858). (Revised by BJ, 23-Jun-2019.) Definitial form. (Revised by Wolf Lammen, 23-Feb-2025.)

Assertion

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

Proof of Theorem abbib

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfcleq 2723	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		nfsab1 2716	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜑}
3		nfsab1 2716	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜓}
4	2, 3	nfbi 1903	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓})
5		nfv 1914	. . 3 ⊢ Ⅎ𝑦(𝜑 ↔ 𝜓)
6		df-clab 2709	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
7		sbequ12r 2253	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
8	6, 7	bitrid 283	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))
9		df-clab 2709	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
10		sbequ12r 2253	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜓 ↔ 𝜓))
11	9, 10	bitrid 283	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓))
12	8, 11	bibi12d 345	. . 3 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜑 ↔ 𝜓)))
13	4, 5, 12	cbvalv1 2339	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑥(𝜑 ↔ 𝜓))
14	1, 13	bitri 275	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

Syntax hints:   ↔ wb 206  ∀wal 1538   = wceq 1540  [wsb 2065   ∈ wcel 2109  {cab 2708

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2709  df-cleq 2722

This theorem is referenced by:  eqabb  2868  nabbib  3029  rabbi  3439  ab0  4346  absn  4612  karden  9855  abeqabi  43404  elnev  44434  csbingVD  44880  csbsngVD  44889  csbxpgVD  44890  csbrngVD  44892  csbunigVD  44894  csbfv12gALTVD  44895