Theorem ercl2 8724

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 8723	. 2 ⊢ (𝜑 → 𝐵𝑅𝐴)
4	1, 3	ercl 8722	1 ⊢ (𝜑 → 𝐵 ∈ 𝑋)

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2145   class class class wbr 5103   Er wer 8707

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733  ax-sep 5249  ax-pr 5391

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 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-xp 5657  df-rel 5658  df-cnv 5659  df-dm 5661  df-er 8710

This theorem is used by:  qliftfun  8816  efgcpbl2  19964  frgpcpbl  19966