Theorem seeq1 5588

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 3974	. . 3 ⊢ (𝑅 = 𝑆 → 𝑆 ⊆ 𝑅)
2		sess1 5583	. . 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 3973	. . 3 ⊢ (𝑅 = 𝑆 → 𝑅 ⊆ 𝑆)
5		sess1 5583	. . 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 213	1 ⊢ (𝑅 = 𝑆 → (𝑅 Se 𝐴 ↔ 𝑆 Se 𝐴))

Syntax hints:   → wi 4   ↔ wb 207   = wceq 1547   ⊆ wss 3883   Se wse 5569

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711  ax-sep 5218

This theorem depends on definitions:  df-bi 208  df-an 397  df-3an 1094  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-ral 3054  df-rab 3392  df-v 3433  df-in 3890  df-ss 3900  df-br 5073  df-se 5572

This theorem is referenced by:  seeq12d  5590  oieq1  9417