Theorem ercl2 8738

Description: Elementhood in the field of an equivalence relation. (Contributed by Mario Carneiro, 12-Aug-2015.)

Hypotheses

Ref	Expression
ersym.1	⊢ (𝜑 → 𝑅 Er 𝑋)
ersym.2	⊢ (𝜑 → 𝐴𝑅𝐵)

Assertion

Ref	Expression
ercl2	⊢ (𝜑 → 𝐵 ∈ 𝑋)

Proof of Theorem ercl2

Step	Hyp	Ref	Expression
1		ersym.1	. 2 ⊢ (𝜑 → 𝑅 Er 𝑋)
2		ersym.2	. . 3 ⊢ (𝜑 → 𝐴𝑅𝐵)
3	1, 2	ersym 8737	. 2 ⊢ (𝜑 → 𝐵𝑅𝐴)
4	1, 3	ercl 8736	1 ⊢ (𝜑 → 𝐵 ∈ 𝑋)

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2144   class class class wbr 5102   Er wer 8721

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  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-xp 5657  df-rel 5658  df-cnv 5659  df-dm 5661  df-er 8724

This theorem is used by:  qliftfun  8830  efgcpbl2  19980  frgpcpbl  19982