Theorem heeq1 44884

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 2760	. 2 ⊢ 𝐴 = 𝐴
2		heeq12 44883	. 2 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐴) → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐴))
3	1, 2	mpan2 704	1 ⊢ (𝑅 = 𝑆 → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569   hereditary whe 44879

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-xp 5656  df-cnv 5658  df-dm 5660  df-rn 5661  df-res 5662  df-ima 5663  df-he 44880

This theorem is used by:  0heALT  44890