Theorem seeq2 5651

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 4039	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		sess2 5647	. . 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 4038	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		sess2 5647	. . 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 1534   ⊆ wss 3947   Se wse 5631

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2699  ax-sep 5299

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-nfc 2881  df-ral 3059  df-rab 3430  df-v 3473  df-in 3954  df-ss 3964  df-se 5634

This theorem is referenced by:  oieq2  9537