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Theorem heeq12 41384
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 483 . . . 4 ((𝑅 = 𝑆𝐴 = 𝐵) → 𝑅 = 𝑆)
2 simpr 485 . . . 4 ((𝑅 = 𝑆𝐴 = 𝐵) → 𝐴 = 𝐵)
31, 2imaeq12d 5970 . . 3 ((𝑅 = 𝑆𝐴 = 𝐵) → (𝑅𝐴) = (𝑆𝐵))
43, 2sseq12d 3954 . 2 ((𝑅 = 𝑆𝐴 = 𝐵) → ((𝑅𝐴) ⊆ 𝐴 ↔ (𝑆𝐵) ⊆ 𝐵))
5 df-he 41381 . 2 (𝑅 hereditary 𝐴 ↔ (𝑅𝐴) ⊆ 𝐴)
6 df-he 41381 . 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 205  wa 396   = wceq 1539  wss 3887  cima 5592   hereditary whe 41380
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-br 5075  df-opab 5137  df-xp 5595  df-cnv 5597  df-dm 5599  df-rn 5600  df-res 5601  df-ima 5602  df-he 41381
This theorem is referenced by:  heeq1  41385  heeq2  41386  frege77  41548
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