Theorem seeq2 5621

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

Assertion

Ref	Expression
seeq2	⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵))

Proof of Theorem seeq2

Step	Hyp	Ref	Expression
1		eqimss2 3967	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		sess2 5616	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Se 𝐴 → 𝑅 Se 𝐵))
3	1, 2	syl 18	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 → 𝑅 Se 𝐵))
4		eqimss 3966	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		sess2 5616	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Se 𝐵 → 𝑅 Se 𝐴))
6	4, 5	syl 18	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐵 → 𝑅 Se 𝐴))
7	3, 6	impbid 215	1 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569   ⊆ wss 3895   Se wse 5601

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rab 3413  df-v 3452  df-ss 3896  df-in 4091  df-se 5604

This theorem is used by:  seeq12d  5622  oieq2  9510