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Theorem bj-indeq 10053
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 df-bj-ind 10051 . 2 (Ind 𝐴 ↔ (∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴))
2 df-bj-ind 10051 . . 3 (Ind 𝐵 ↔ (∅ ∈ 𝐵 ∧ ∀𝑥𝐵 suc 𝑥𝐵))
3 eleq2 2101 . . . . 5 (𝐴 = 𝐵 → (∅ ∈ 𝐴 ↔ ∅ ∈ 𝐵))
43bicomd 129 . . . 4 (𝐴 = 𝐵 → (∅ ∈ 𝐵 ↔ ∅ ∈ 𝐴))
5 eleq2 2101 . . . . . 6 (𝐴 = 𝐵 → (suc 𝑥𝐴 ↔ suc 𝑥𝐵))
65raleqbi1dv 2513 . . . . 5 (𝐴 = 𝐵 → (∀𝑥𝐴 suc 𝑥𝐴 ↔ ∀𝑥𝐵 suc 𝑥𝐵))
76bicomd 129 . . . 4 (𝐴 = 𝐵 → (∀𝑥𝐵 suc 𝑥𝐵 ↔ ∀𝑥𝐴 suc 𝑥𝐴))
84, 7anbi12d 442 . . 3 (𝐴 = 𝐵 → ((∅ ∈ 𝐵 ∧ ∀𝑥𝐵 suc 𝑥𝐵) ↔ (∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴)))
92, 8syl5rbb 182 . 2 (𝐴 = 𝐵 → ((∅ ∈ 𝐴 ∧ ∀𝑥𝐴 suc 𝑥𝐴) ↔ Ind 𝐵))
101, 9syl5bb 181 1 (𝐴 = 𝐵 → (Ind 𝐴 ↔ Ind 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 97  wb 98   = wceq 1243  wcel 1393  wral 2306  c0 3224  suc csuc 4102  Ind wind 10050
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-bj-ind 10051
This theorem is referenced by:  bj-omind  10058  bj-omssind  10059  bj-ssom  10060  bj-om  10061  bj-2inf  10062  peano5setOLD  10065
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