Theorem heeq1 43728

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

Assertion

Ref	Expression
heeq1	⊢ (𝑅 = 𝑆 → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐴))

Proof of Theorem heeq1

Step	Hyp	Ref	Expression
1		eqid 2734	. 2 ⊢ 𝐴 = 𝐴
2		heeq12 43727	. 2 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐴) → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐴))
3	1, 2	mpan2 691	1 ⊢ (𝑅 = 𝑆 → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐴))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1539   hereditary whe 43723

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1779  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-rab 3420  df-v 3465  df-dif 3934  df-un 3936  df-in 3938  df-ss 3948  df-nul 4314  df-if 4506  df-sn 4607  df-pr 4609  df-op 4613  df-br 5124  df-opab 5186  df-xp 5671  df-cnv 5673  df-dm 5675  df-rn 5676  df-res 5677  df-ima 5678  df-he 43724

This theorem is referenced by:  0heALT  43734