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Theorem bj-indeq 15575
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 2260 . . 3 (𝐴 = 𝐵 → (∅ ∈ 𝐴 ↔ ∅ ∈ 𝐵))
2 eleq2 2260 . . . 4 (𝐴 = 𝐵 → (suc 𝑥𝐴 ↔ suc 𝑥𝐵))
32raleqbi1dv 2705 . . 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 15573 . 2 (Ind 𝐴 ↔ (∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴))
6 df-bj-ind 15573 . 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 1364  wcel 2167  wral 2475  c0 3450  suc csuc 4400  Ind wind 15572
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 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178
This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ral 2480  df-bj-ind 15573
This theorem is referenced by:  bj-omind  15580  bj-omssind  15581  bj-ssom  15582  bj-om  15583  bj-2inf  15584
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