Theorem abpr 43422

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 4646	. 2 ⊢ {𝑌, 𝑍} = {𝑥 ∣ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)}
2	1	abeqabi 43421	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌, 𝑍} ↔ ∀𝑥(𝜑 ↔ (𝑥 = 𝑌 ∨ 𝑥 = 𝑍)))

Syntax hints:   ↔ wb 206   ∨ wo 848  ∀wal 1538   = wceq 1540  {cab 2714  {cpr 4628

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-v 3482  df-un 3956  df-sn 4627  df-pr 4629

This theorem is referenced by: (None)