Theorem heeq1 41247

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 2739	. 2 ⊢ 𝐴 = 𝐴
2		heeq12 41246	. 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 209   = wceq 1543   hereditary whe 41242

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-8 2114  ax-9 2122  ax-ext 2710

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-3an 1091  df-tru 1546  df-fal 1556  df-ex 1788  df-sb 2073  df-clab 2717  df-cleq 2731  df-clel 2818  df-rab 3073  df-v 3425  df-dif 3887  df-un 3889  df-in 3891  df-ss 3901  df-nul 4255  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-opab 5133  df-xp 5585  df-cnv 5587  df-dm 5589  df-rn 5590  df-res 5591  df-ima 5592  df-he 41243

This theorem is referenced by:  0heALT  41253