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Theorem heeq12 43772
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 6035 . . 3 ((𝑅 = 𝑆𝐴 = 𝐵) → (𝑅𝐴) = (𝑆𝐵))
43, 2sseq12d 3983 . 2 ((𝑅 = 𝑆𝐴 = 𝐵) → ((𝑅𝐴) ⊆ 𝐴 ↔ (𝑆𝐵) ⊆ 𝐵))
5 df-he 43769 . 2 (𝑅 hereditary 𝐴 ↔ (𝑅𝐴) ⊆ 𝐴)
6 df-he 43769 . 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 3917  cima 5644   hereditary whe 43768
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 2008  ax-8 2111  ax-9 2119  ax-ext 2702
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-in 3924  df-ss 3934  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-br 5111  df-opab 5173  df-xp 5647  df-cnv 5649  df-dm 5651  df-rn 5652  df-res 5653  df-ima 5654  df-he 43769
This theorem is referenced by:  heeq1  43773  heeq2  43774  frege77  43936
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