Theorem elprg 1822

Description: A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15, generalized.

Assertion

Ref	Expression
elprg	⊢ (A ∈ D → (A ∈ {B, C} ↔ (A = B ∨ A = C)))

Proof of Theorem elprg

Step	Hyp	Ref	Expression
1		cleq1 1107	. . 3 ⊢ (x = A → (x = B ↔ A = B))
2		cleq1 1107	. . 3 ⊢ (x = A → (x = C ↔ A = C))
3	1, 2	orbi12d 475	. 2 ⊢ (x = A → ((x = B ∨ x = C) ↔ (A = B ∨ A = C)))
4		dfpr2 1821	. 2 ⊢ {B, C} = {x∣(x = B ∨ x = C)}
5	3, 4	elab2g 1418	1 ⊢ (A ∈ D → (A ∈ {B, C} ↔ (A = B ∨ A = C)))

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195   = wceq 1091   ∈ wcel 1092  {cpr 1809

This theorem is referenced by:  elpr 1823  elsncg 1825  snsspr 1853

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-sn 1811  df-pr 1812

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