Theorem dfpr2 4605

Description: Alternate definition of a pair. Definition 5.1 of [TakeutiZaring] p. 15. (Contributed by NM, 24-Apr-1994.)

Assertion

Ref	Expression
dfpr2	⊢ {𝐴, 𝐵} = {𝑥 ∣ (𝑥 = 𝐴 ∨ 𝑥 = 𝐵)}

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem dfpr2

Step	Hyp	Ref	Expression
1		df-pr 4591	. 2 ⊢ {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
2		elun 4155	. . . 4 ⊢ (𝑥 ∈ ({𝐴} ∪ {𝐵}) ↔ (𝑥 ∈ {𝐴} ∨ 𝑥 ∈ {𝐵}))
3		velsn 4600	. . . . 5 ⊢ (𝑥 ∈ {𝐴} ↔ 𝑥 = 𝐴)
4		velsn 4600	. . . . 5 ⊢ (𝑥 ∈ {𝐵} ↔ 𝑥 = 𝐵)
5	3, 4	orbi12i 928	. . . 4 ⊢ ((𝑥 ∈ {𝐴} ∨ 𝑥 ∈ {𝐵}) ↔ (𝑥 = 𝐴 ∨ 𝑥 = 𝐵))
6	2, 5	bitri 278	. . 3 ⊢ (𝑥 ∈ ({𝐴} ∪ {𝐵}) ↔ (𝑥 = 𝐴 ∨ 𝑥 = 𝐵))
7	6	eqabi 2899	. 2 ⊢ ({𝐴} ∪ {𝐵}) = {𝑥 ∣ (𝑥 = 𝐴 ∨ 𝑥 = 𝐵)}
8	1, 7	eqtri 2786	1 ⊢ {𝐴, 𝐵} = {𝑥 ∣ (𝑥 = 𝐴 ∨ 𝑥 = 𝐵)}

This proof depends on syntax axioms:   ∨ wo 861   = wceq 1570   ∈ wcel 2147  {cab 2741   ∪ cun 4152  {csn 4588  {cpr 4590

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

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-un 4154  df-sn 4589  df-pr 4591

This theorem is used by:  dfsn2ALT  4606  elprg  4607  nfpr  4653  pwsn  4830  pwpw0  4834  zfpair  5387  grothprimlem  10945  nb3grprlem1  29991  rabsspr  33169  abpr  44602