Theorem seeq2 5585

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 3989	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		sess2 5580	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Se 𝐴 → 𝑅 Se 𝐵))
3	1, 2	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 → 𝑅 Se 𝐵))
4		eqimss 3988	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		sess2 5580	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Se 𝐵 → 𝑅 Se 𝐴))
6	4, 5	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐵 → 𝑅 Se 𝐴))
7	3, 6	impbid 212	1 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1541   ⊆ wss 3897   Se wse 5565

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703  ax-sep 5232

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ral 3048  df-rab 3396  df-v 3438  df-in 3904  df-ss 3914  df-se 5568

This theorem is referenced by:  seeq12d  5586  oieq2  9399