Theorem heeq1 41355

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 2740	. 2 ⊢ 𝐴 = 𝐴
2		heeq12 41354	. 2 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐴) → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐴))
3	1, 2	mpan2 688	1 ⊢ (𝑅 = 𝑆 → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐴))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1542   hereditary whe 41350

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-8 2112  ax-9 2120  ax-ext 2711

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1545  df-fal 1555  df-ex 1787  df-sb 2072  df-clab 2718  df-cleq 2732  df-clel 2818  df-rab 3075  df-v 3433  df-dif 3895  df-un 3897  df-in 3899  df-ss 3909  df-nul 4263  df-if 4466  df-sn 4568  df-pr 4570  df-op 4574  df-br 5080  df-opab 5142  df-xp 5596  df-cnv 5598  df-dm 5600  df-rn 5601  df-res 5602  df-ima 5603  df-he 41351

This theorem is referenced by:  0heALT  41361