Theorem abbib 2832

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 2149 and df-clel 2839 (by avoiding use of cleqh 2893). (Revised by BJ, 23-Jun-2019.) Reverse the biconditional. (Revised by Wolf Lammen, 23-Feb-2025.) (Proof shortened by SN, 22-Oct-2026.)

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		df-clab 2742	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
2		df-clab 2742	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
3	1, 2	bibi12i 342	. . . 4 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))
4		sbbi 2344	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))
5	3, 4	bitr4i 281	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ [𝑦 / 𝑥](𝜑 ↔ 𝜓))
6	5	albii 1852	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑦[𝑦 / 𝑥](𝜑 ↔ 𝜓))
7		dfcleq 2756	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
8		sb8v 2385	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ ∀𝑦[𝑦 / 𝑥](𝜑 ↔ 𝜓))
9	6, 7, 8	3bitr4i 306	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568   = wceq 1570  [wsb 2101   ∈ wcel 2147  {cab 2741

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755

This theorem is used by:  eqabb  2903  nabbib  3064  rabbi  3445  ab0  4332  absn  4604  kardenOLD  9990  abeqabi  44612  elnev  45623  csbingVD  46068  csbsngVD  46077  csbxpgVD  46078  csbrngVD  46080  csbunigVD  46082  csbfv12gALTVD  46083