Theorem absn 4386

Description: Condition for a class abstraction to be a singleton. Formerly part of proof of dfiota2 6065. (Contributed by Andrew Salmon, 30-Jun-2011.) (Revised by AV, 24-Aug-2022.)

Assertion

Ref	Expression
absn	⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))

Distinct variable group:   𝑥,𝑌

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem absn

Step	Hyp	Ref	Expression
1		df-sn 4369	. . 3 ⊢ {𝑌} = {𝑥 ∣ 𝑥 = 𝑌}
2	1	eqeq2i 2811	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌})
3		abbi 2914	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝑥 = 𝑌) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌})
4	2, 3	bitr4i 270	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))

Syntax hints:   ↔ wb 198  ∀wal 1651   = wceq 1653  {cab 2785  {csn 4368

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-ext 2777

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-sn 4369

This theorem is referenced by:  rabeqsn  4405  dfiota2  6065  dfaiota2  41935  aiotaval  41942