Users' Mathboxes Mathbox for BJ < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >   Mathboxes  >  bj-indeq GIF version

Theorem bj-indeq 15865
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 2269 . . 3 (𝐴 = 𝐵 → (∅ ∈ 𝐴 ↔ ∅ ∈ 𝐵))
2 eleq2 2269 . . . 4 (𝐴 = 𝐵 → (suc 𝑥𝐴 ↔ suc 𝑥𝐵))
32raleqbi1dv 2714 . . 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 15863 . 2 (Ind 𝐴 ↔ (∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴))
6 df-bj-ind 15863 . 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 2176  wral 2484  c0 3460  suc csuc 4412  Ind wind 15862
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 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187
This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-cleq 2198  df-clel 2201  df-nfc 2337  df-ral 2489  df-bj-ind 15863
This theorem is referenced by:  bj-omind  15870  bj-omssind  15871  bj-ssom  15872  bj-om  15873  bj-2inf  15874
  Copyright terms: Public domain W3C validator