Theorem heeq12 44535

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

Assertion

Ref	Expression
heeq12	⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐵))

Proof of Theorem heeq12

Step	Hyp	Ref	Expression
1		simpl 488	. . . 4 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → 𝑅 = 𝑆)
2		simpr 490	. . . 4 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵)
3	1, 2	imaeq12d 6065	. . 3 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → (𝑅 “ 𝐴) = (𝑆 “ 𝐵))
4	3, 2	sseq12d 3971	. 2 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → ((𝑅 “ 𝐴) ⊆ 𝐴 ↔ (𝑆 “ 𝐵) ⊆ 𝐵))
5		df-he 44532	. 2 ⊢ (𝑅 hereditary 𝐴 ↔ (𝑅 “ 𝐴) ⊆ 𝐴)
6		df-he 44532	. 2 ⊢ (𝑆 hereditary 𝐵 ↔ (𝑆 “ 𝐵) ⊆ 𝐵)
7	4, 5, 6	3bitr4g 317	1 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → (𝑅 hereditary 𝐴 ↔ 𝑆 hereditary 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ⊆ wss 3906   “ cima 5666   hereditary whe 44531

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 2737

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 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-xp 5669  df-cnv 5671  df-dm 5673  df-rn 5674  df-res 5675  df-ima 5676  df-he 44532

This theorem is used by:  heeq1  44536  heeq2  44537  frege77  44699