Theorem abpr 44550

Description: Condition for a class abstraction to be a pair. (Contributed by RP, 25-Aug-2024.)

Assertion

Ref	Expression
abpr	⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))

Distinct variable groups:   𝑥,𝑌   𝑥,𝑍

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem abpr

Step	Hyp	Ref	Expression
1		dfpr2 4604	. 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)}
2	1	abeqabi 44549	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861  ∀wal 1567   = wceq 1569  {cab 2741  {cpr 4589

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2196  ax-12 2215  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3455  df-un 4153  df-sn 4588  df-pr 4590

This theorem is used by: (None)