Theorem heeq2 41386

Description: Equality law for relations being herditary over a class. (Contributed by RP, 27-Mar-2020.)

Assertion

Ref	Expression
heeq2	⊢ (𝐴 = 𝐵 → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))

Proof of Theorem heeq2

Step	Hyp	Ref	Expression
1		eqid 2738	. 2 ⊢ 𝑅 = 𝑅
2		heeq12 41384	. 2 ⊢ ((𝑅 = 𝑅 ∧ 𝐴 = 𝐵) → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))
3	1, 2	mpan 687	1 ⊢ (𝐴 = 𝐵 → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1539   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: (None)