Theorem seeq1 5508

Description: Equality theorem for the set-like predicate. (Contributed by Mario Carneiro, 24-Jun-2015.)

Assertion

Ref	Expression
seeq1	⊢ (𝑅 = 𝑆 → (𝑅 Se 𝐴 ↔ 𝑆 Se 𝐴))

Proof of Theorem seeq1

Step	Hyp	Ref	Expression
1		eqimss2 3944	. . 3 ⊢ (𝑅 = 𝑆 → 𝑆 ⊆ 𝑅)
2		sess1 5504	. . 3 ⊢ (𝑆 ⊆ 𝑅 → (𝑅 Se 𝐴 → 𝑆 Se 𝐴))
3	1, 2	syl 17	. 2 ⊢ (𝑅 = 𝑆 → (𝑅 Se 𝐴 → 𝑆 Se 𝐴))
4		eqimss 3943	. . 3 ⊢ (𝑅 = 𝑆 → 𝑅 ⊆ 𝑆)
5		sess1 5504	. . 3 ⊢ (𝑅 ⊆ 𝑆 → (𝑆 Se 𝐴 → 𝑅 Se 𝐴))
6	4, 5	syl 17	. 2 ⊢ (𝑅 = 𝑆 → (𝑆 Se 𝐴 → 𝑅 Se 𝐴))
7	3, 6	impbid 215	1 ⊢ (𝑅 = 𝑆 → (𝑅 Se 𝐴 ↔ 𝑆 Se 𝐴))

Syntax hints:   → wi 4   ↔ wb 209   = wceq 1543   ⊆ wss 3853   Se wse 5492

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-10 2143  ax-11 2160  ax-12 2177  ax-ext 2708  ax-sep 5177

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-tru 1546  df-ex 1788  df-nf 1792  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809  df-nfc 2879  df-ral 3056  df-rab 3060  df-v 3400  df-in 3860  df-ss 3870  df-br 5040  df-se 5495

This theorem is referenced by:  oieq1  9106