Theorem ercl2 8769

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 8768	. 2 ⊢ (𝜑 → 𝐵𝑅𝐴)
4	1, 3	ercl 8767	1 ⊢ (𝜑 → 𝐵 ∈ 𝑋)

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2147   class class class wbr 5108   Er wer 8752

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5255  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-br 5109  df-opab 5174  df-xp 5664  df-rel 5665  df-cnv 5666  df-dm 5668  df-er 8755

This theorem is used by:  qliftfun  8861  efgcpbl2  20022  frgpcpbl  20024  ecscottd  35885