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Theorem abbib 2800
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 2101 and df-clel 2806 (by avoiding use of cleqh 2859). (Revised by BJ, 23-Jun-2019.) Definitial form. (Revised by Wolf Lammen, 23-Feb-2025.)
Assertion
Ref Expression
abbib ({𝑥𝜑} = {𝑥𝜓} ↔ ∀𝑥(𝜑𝜓))

Proof of Theorem abbib
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 dfcleq 2721 . 2 ({𝑥𝜑} = {𝑥𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓}))
2 nfsab1 2713 . . . 4 𝑥 𝑦 ∈ {𝑥𝜑}
3 nfsab1 2713 . . . 4 𝑥 𝑦 ∈ {𝑥𝜓}
42, 3nfbi 1899 . . 3 𝑥(𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓})
5 nfv 1910 . . 3 𝑦(𝜑𝜓)
6 df-clab 2706 . . . . 5 (𝑦 ∈ {𝑥𝜑} ↔ [𝑦 / 𝑥]𝜑)
7 sbequ12r 2240 . . . . 5 (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑𝜑))
86, 7bitrid 283 . . . 4 (𝑦 = 𝑥 → (𝑦 ∈ {𝑥𝜑} ↔ 𝜑))
9 df-clab 2706 . . . . 5 (𝑦 ∈ {𝑥𝜓} ↔ [𝑦 / 𝑥]𝜓)
10 sbequ12r 2240 . . . . 5 (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜓𝜓))
119, 10bitrid 283 . . . 4 (𝑦 = 𝑥 → (𝑦 ∈ {𝑥𝜓} ↔ 𝜓))
128, 11bibi12d 345 . . 3 (𝑦 = 𝑥 → ((𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓}) ↔ (𝜑𝜓)))
134, 5, 12cbvalv1 2333 . 2 (∀𝑦(𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓}) ↔ ∀𝑥(𝜑𝜓))
141, 13bitri 275 1 ({𝑥𝜑} = {𝑥𝜓} ↔ ∀𝑥(𝜑𝜓))
Colors of variables: wff setvar class
Syntax hints:  wb 205  wal 1532   = wceq 1534  [wsb 2060  wcel 2099  {cab 2705
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2699
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2706  df-cleq 2720
This theorem is referenced by:  eqabb  2869  nabbib  3042  rabbi  3459  ab0  4375  absn  4647  karden  9918  abeqabi  42838  elnev  43875  csbingVD  44323  csbsngVD  44332  csbxpgVD  44333  csbrngVD  44335  csbunigVD  44337  csbfv12gALTVD  44338
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