Theorem seeq2 5637

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 3999	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		sess2 5632	. . 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 3998	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		sess2 5632	. . 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 1570   ⊆ wss 3908   Se wse 5617

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738  ax-sep 5262

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-rab 3420  df-v 3460  df-in 3915  df-ss 3925  df-se 5620

This theorem is used by:  seeq12d  5638  oieq2  9485