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Theorem bj-indeq 16292
Description: Equality property for Ind. (Contributed by BJ, 30-Nov-2019.)
Assertion
Ref Expression
bj-indeq (𝐴 = 𝐵 → (Ind 𝐴 ↔ Ind 𝐵))

Proof of Theorem bj-indeq
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 eleq2 2293 . . 3 (𝐴 = 𝐵 → (∅ ∈ 𝐴 ↔ ∅ ∈ 𝐵))
2 eleq2 2293 . . . 4 (𝐴 = 𝐵 → (suc 𝑥𝐴 ↔ suc 𝑥𝐵))
32raleqbi1dv 2740 . . 3 (𝐴 = 𝐵 → (∀𝑥𝐴 suc 𝑥𝐴 ↔ ∀𝑥𝐵 suc 𝑥𝐵))
41, 3anbi12d 473 . 2 (𝐴 = 𝐵 → ((∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴) ↔ (∅ ∈ 𝐵 ∧ ∀𝑥𝐵 suc 𝑥𝐵)))
5 df-bj-ind 16290 . 2 (Ind 𝐴 ↔ (∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴))
6 df-bj-ind 16290 . 2 (Ind 𝐵 ↔ (∅ ∈ 𝐵 ∧ ∀𝑥𝐵 suc 𝑥𝐵))
74, 5, 63bitr4g 223 1 (𝐴 = 𝐵 → (Ind 𝐴 ↔ Ind 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105   = wceq 1395  wcel 2200  wral 2508  c0 3491  suc csuc 4456  Ind wind 16289
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211
This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-bj-ind 16290
This theorem is referenced by:  bj-omind  16297  bj-omssind  16298  bj-ssom  16299  bj-om  16300  bj-2inf  16301
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