Theorem heeq2 44545

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

Assertion

Ref	Expression
heeq2	⊢ (𝐴 = 𝐵 → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))

Proof of Theorem heeq2

Step	Hyp	Ref	Expression
1		eqid 2766	. 2 ⊢ 𝑅 = 𝑅
2		heeq12 44543	. 2 ⊢ ((𝑅 = 𝑅 ∧ 𝐴 = 𝐵) → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))
3	1, 2	mpan 703	1 ⊢ (𝐴 = 𝐵 → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   hereditary whe 44539

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5115  df-opab 5179  df-xp 5672  df-cnv 5674  df-dm 5676  df-rn 5677  df-res 5678  df-ima 5679  df-he 44540

This theorem is used by: (None)