Theorem abbib 2830

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 2143 and df-clel 2836 (by avoiding use of cleqh 2890). (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 2754	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		nfsab1 2747	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜑}
3		nfsab1 2747	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜓}
4	2, 3	nfbi 1922	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓})
5		nfv 1933	. . 3 ⊢ Ⅎ𝑦(𝜑 ↔ 𝜓)
6		df-clab 2740	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
7		sbequ12r 2286	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
8	6, 7	bitrid 285	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))
9		df-clab 2740	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
10		sbequ12r 2286	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜓 ↔ 𝜓))
11	9, 10	bitrid 285	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓))
12	8, 11	bibi12d 347	. . 3 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜑 ↔ 𝜓)))
13	4, 5, 12	cbvalv1 2371	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑥(𝜑 ↔ 𝜓))
14	1, 13	bitri 277	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

Syntax hints:   ↔ wb 208  ∀wal 1557   = wceq 1559  [wsb 2089   ∈ wcel 2141  {cab 2739

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753

This theorem is referenced by:  eqabb  2900  nabbib  3059  rabbi  3443  ab0  4330  absn  4599  karden  9846  abeqabi  43944  elnev  44973  csbingVD  45419  csbsngVD  45428  csbxpgVD  45429  csbrngVD  45431  csbunigVD  45433  csbfv12gALTVD  45434