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Theorem bj-indeq 16064
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 2271 . . 3 (𝐴 = 𝐵 → (∅ ∈ 𝐴 ↔ ∅ ∈ 𝐵))
2 eleq2 2271 . . . 4 (𝐴 = 𝐵 → (suc 𝑥𝐴 ↔ suc 𝑥𝐵))
32raleqbi1dv 2717 . . 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 16062 . 2 (Ind 𝐴 ↔ (∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴))
6 df-bj-ind 16062 . 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 1373  wcel 2178  wral 2486  c0 3468  suc csuc 4430  Ind wind 16061
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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189
This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-cleq 2200  df-clel 2203  df-nfc 2339  df-ral 2491  df-bj-ind 16062
This theorem is referenced by:  bj-omind  16069  bj-omssind  16070  bj-ssom  16071  bj-om  16072  bj-2inf  16073
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