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Theorem abbib 2796
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 2100 and df-clel 2802 (by avoiding use of cleqh 2855). (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 2717 . 2 ({𝑥𝜑} = {𝑥𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓}))
2 nfsab1 2709 . . . 4 𝑥 𝑦 ∈ {𝑥𝜑}
3 nfsab1 2709 . . . 4 𝑥 𝑦 ∈ {𝑥𝜓}
42, 3nfbi 1898 . . 3 𝑥(𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓})
5 nfv 1909 . . 3 𝑦(𝜑𝜓)
6 df-clab 2702 . . . . 5 (𝑦 ∈ {𝑥𝜑} ↔ [𝑦 / 𝑥]𝜑)
7 sbequ12r 2236 . . . . 5 (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑𝜑))
86, 7bitrid 283 . . . 4 (𝑦 = 𝑥 → (𝑦 ∈ {𝑥𝜑} ↔ 𝜑))
9 df-clab 2702 . . . . 5 (𝑦 ∈ {𝑥𝜓} ↔ [𝑦 / 𝑥]𝜓)
10 sbequ12r 2236 . . . . 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 2329 . 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 1531   = wceq 1533  [wsb 2059  wcel 2098  {cab 2701
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2695
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2702  df-cleq 2716
This theorem is referenced by:  eqabb  2865  nabbib  3037  rabbi  3454  ab0  4367  absn  4639  karden  9887  abeqabi  42708  elnev  43746  csbingVD  44194  csbsngVD  44203  csbxpgVD  44204  csbrngVD  44206  csbunigVD  44208  csbfv12gALTVD  44209
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