Theorem seeq2 5553

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 3974	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		sess2 5549	. . 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 3973	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		sess2 5549	. . 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 211	1 ⊢ (𝐴 = 𝐵 → (𝑅 Se 𝐴 ↔ 𝑅 Se 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1539   ⊆ wss 3883   Se wse 5533

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ral 3068  df-rab 3072  df-v 3424  df-in 3890  df-ss 3900  df-se 5536

This theorem is referenced by:  oieq2  9202