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Theorem abbib 2809
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 2108 and df-clel 2814 (by avoiding use of cleqh 2869). (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 2728 . 2 ({𝑥𝜑} = {𝑥𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓}))
2 nfsab1 2720 . . . 4 𝑥 𝑦 ∈ {𝑥𝜑}
3 nfsab1 2720 . . . 4 𝑥 𝑦 ∈ {𝑥𝜓}
42, 3nfbi 1901 . . 3 𝑥(𝑦 ∈ {𝑥𝜑} ↔ 𝑦 ∈ {𝑥𝜓})
5 nfv 1912 . . 3 𝑦(𝜑𝜓)
6 df-clab 2713 . . . . 5 (𝑦 ∈ {𝑥𝜑} ↔ [𝑦 / 𝑥]𝜑)
7 sbequ12r 2250 . . . . 5 (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑𝜑))
86, 7bitrid 283 . . . 4 (𝑦 = 𝑥 → (𝑦 ∈ {𝑥𝜑} ↔ 𝜑))
9 df-clab 2713 . . . . 5 (𝑦 ∈ {𝑥𝜓} ↔ [𝑦 / 𝑥]𝜓)
10 sbequ12r 2250 . . . . 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 2342 . 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 206  wal 1535   = wceq 1537  [wsb 2062  wcel 2106  {cab 2712
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-ex 1777  df-nf 1781  df-sb 2063  df-clab 2713  df-cleq 2727
This theorem is referenced by:  eqabb  2879  nabbib  3043  rabbi  3465  ab0  4386  absn  4650  karden  9933  abeqabi  43398  elnev  44434  csbingVD  44882  csbsngVD  44891  csbxpgVD  44892  csbrngVD  44894  csbunigVD  44896  csbfv12gALTVD  44897
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