Theorem heeq1 43738

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 2730	. 2 ⊢ 𝐴 = 𝐴
2		heeq12 43737	. 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 1540   hereditary whe 43733

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-rab 3412  df-v 3457  df-dif 3925  df-un 3927  df-in 3929  df-ss 3939  df-nul 4305  df-if 4497  df-sn 4598  df-pr 4600  df-op 4604  df-br 5116  df-opab 5178  df-xp 5652  df-cnv 5654  df-dm 5656  df-rn 5657  df-res 5658  df-ima 5659  df-he 43734

This theorem is referenced by:  0heALT  43744