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Theorem heeq12 43789
Description: Equality law for relations being herditary over a class. (Contributed by RP, 27-Mar-2020.)
Assertion
Ref Expression
heeq12 ((𝑅 = 𝑆𝐴 = 𝐵) → (𝑅 hereditary 𝐴𝑆 hereditary 𝐵))

Proof of Theorem heeq12
StepHypRef Expression
1 simpl 482 . . . 4 ((𝑅 = 𝑆𝐴 = 𝐵) → 𝑅 = 𝑆)
2 simpr 484 . . . 4 ((𝑅 = 𝑆𝐴 = 𝐵) → 𝐴 = 𝐵)
31, 2imaeq12d 6079 . . 3 ((𝑅 = 𝑆𝐴 = 𝐵) → (𝑅𝐴) = (𝑆𝐵))
43, 2sseq12d 4017 . 2 ((𝑅 = 𝑆𝐴 = 𝐵) → ((𝑅𝐴) ⊆ 𝐴 ↔ (𝑆𝐵) ⊆ 𝐵))
5 df-he 43786 . 2 (𝑅 hereditary 𝐴 ↔ (𝑅𝐴) ⊆ 𝐴)
6 df-he 43786 . 2 (𝑆 hereditary 𝐵 ↔ (𝑆𝐵) ⊆ 𝐵)
74, 5, 63bitr4g 314 1 ((𝑅 = 𝑆𝐴 = 𝐵) → (𝑅 hereditary 𝐴𝑆 hereditary 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1540  wss 3951  cima 5688   hereditary whe 43785
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-xp 5691  df-cnv 5693  df-dm 5695  df-rn 5696  df-res 5697  df-ima 5698  df-he 43786
This theorem is referenced by:  heeq1  43790  heeq2  43791  frege77  43953
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