Theorem seeq1 5625

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 3972	. . 3 ⊢ (𝑅 = 𝑆 → 𝑆 ⊆ 𝑅)
2		sess1 5620	. . 3 ⊢ (𝑆 ⊆ 𝑅 → (𝑅 Se 𝐴 → 𝑆 Se 𝐴))
3	1, 2	syl 18	. 2 ⊢ (𝑅 = 𝑆 → (𝑅 Se 𝐴 → 𝑆 Se 𝐴))
4		eqimss 3971	. . 3 ⊢ (𝑅 = 𝑆 → 𝑅 ⊆ 𝑆)
5		sess1 5620	. . 3 ⊢ (𝑅 ⊆ 𝑆 → (𝑆 Se 𝐴 → 𝑅 Se 𝐴))
6	4, 5	syl 18	. 2 ⊢ (𝑅 = 𝑆 → (𝑆 Se 𝐴 → 𝑅 Se 𝐴))
7	3, 6	impbid 215	1 ⊢ (𝑅 = 𝑆 → (𝑅 Se 𝐴 ↔ 𝑆 Se 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ⊆ wss 3900   Se wse 5606

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5252

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rab 3417  df-v 3456  df-ss 3901  df-in 4094  df-br 5107  df-se 5609

This theorem is used by:  seeq12d  5627  oieq1  9530