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Theorem heeq12 40398
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 486 . . . 4 ((𝑅 = 𝑆𝐴 = 𝐵) → 𝑅 = 𝑆)
2 simpr 488 . . . 4 ((𝑅 = 𝑆𝐴 = 𝐵) → 𝐴 = 𝐵)
31, 2imaeq12d 5917 . . 3 ((𝑅 = 𝑆𝐴 = 𝐵) → (𝑅𝐴) = (𝑆𝐵))
43, 2sseq12d 3986 . 2 ((𝑅 = 𝑆𝐴 = 𝐵) → ((𝑅𝐴) ⊆ 𝐴 ↔ (𝑆𝐵) ⊆ 𝐵))
5 df-he 40395 . 2 (𝑅 hereditary 𝐴 ↔ (𝑅𝐴) ⊆ 𝐴)
6 df-he 40395 . 2 (𝑆 hereditary 𝐵 ↔ (𝑆𝐵) ⊆ 𝐵)
74, 5, 63bitr4g 317 1 ((𝑅 = 𝑆𝐴 = 𝐵) → (𝑅 hereditary 𝐴𝑆 hereditary 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 399   = wceq 1538  wss 3919  cima 5545   hereditary whe 40394
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-10 2146  ax-11 2162  ax-12 2179  ax-ext 2796
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2964  df-rab 3142  df-v 3482  df-un 3924  df-in 3926  df-ss 3936  df-sn 4551  df-pr 4553  df-op 4557  df-br 5053  df-opab 5115  df-xp 5548  df-cnv 5550  df-dm 5552  df-rn 5553  df-res 5554  df-ima 5555  df-he 40395
This theorem is referenced by:  heeq1  40399  heeq2  40400  frege77  40562
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