Theorem heeq2 44163

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 2737	. 2 ⊢ 𝑅 = 𝑅
2		heeq12 44161	. 2 ⊢ ((𝑅 = 𝑅 ∧ 𝐴 = 𝐵) → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))
3	1, 2	mpan 691	1 ⊢ (𝐴 = 𝐵 → (𝑅 hereditary 𝐴 ↔ 𝑅 hereditary 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1542   hereditary whe 44157

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-xp 5640  df-cnv 5642  df-dm 5644  df-rn 5645  df-res 5646  df-ima 5647  df-he 44158

This theorem is referenced by: (None)