Theorem abpr 44195

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 4612	. 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)}
2	1	abeqabi 44194	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861  ∀wal 1568   = wceq 1570  {cab 2743  {cpr 4593

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 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-un 3911  df-sn 4592  df-pr 4594

This theorem is used by: (None)