Theorem seeq1 5618

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 3996	. . 3 ⊢ (𝑅 = 𝑆 → 𝑆 ⊆ 𝑅)
2		sess1 5613	. . 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 3995	. . 3 ⊢ (𝑅 = 𝑆 → 𝑅 ⊆ 𝑆)
5		sess1 5613	. . 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 214	1 ⊢ (𝑅 = 𝑆 → (𝑅 Se 𝐴 ↔ 𝑆 Se 𝐴))

Syntax hints:   → wi 4   ↔ wb 208   = wceq 1561   ⊆ wss 3905   Se wse 5599

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1816  ax-4 1830  ax-5 1931  ax-6 1988  ax-7 2029  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5247

This theorem depends on definitions:  df-bi 209  df-an 400  df-3an 1101  df-tru 1564  df-ex 1801  df-sb 2092  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3078  df-rab 3416  df-v 3457  df-in 3912  df-ss 3922  df-br 5102  df-se 5602

This theorem is referenced by:  seeq12d  5620  oieq1  9461