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Theorem heeq1 40001
Description: Equality law for relations being herditary over a class. (Contributed by RP, 27-Mar-2020.)
Assertion
Ref Expression
heeq1 (𝑅 = 𝑆 → (𝑅 hereditary 𝐴𝑆 hereditary 𝐴))

Proof of Theorem heeq1
StepHypRef Expression
1 eqid 2818 . 2 𝐴 = 𝐴
2 heeq12 40000 . 2 ((𝑅 = 𝑆𝐴 = 𝐴) → (𝑅 hereditary 𝐴𝑆 hereditary 𝐴))
31, 2mpan2 687 1 (𝑅 = 𝑆 → (𝑅 hereditary 𝐴𝑆 hereditary 𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207   = wceq 1528   hereditary whe 39996
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-3an 1081  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-rab 3144  df-v 3494  df-dif 3936  df-un 3938  df-in 3940  df-ss 3949  df-nul 4289  df-if 4464  df-sn 4558  df-pr 4560  df-op 4564  df-br 5058  df-opab 5120  df-xp 5554  df-cnv 5556  df-dm 5558  df-rn 5559  df-res 5560  df-ima 5561  df-he 39997
This theorem is referenced by:  0heALT  40007
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