Theorem absn 4646

Description: Condition for a class abstraction to be a singleton. Formerly part of proof of dfiota2 6496. (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 4629	. . 3 ⊢ {𝑌} = {𝑥 ∣ 𝑥 = 𝑌}
2	1	eqeq2i 2745	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌})
3		abbib 2804	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))
4	2, 3	bitri 274	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))

Syntax hints:   ↔ wb 205  ∀wal 1539   = wceq 1541  {cab 2709  {csn 4628

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2710  df-cleq 2724  df-sn 4629

This theorem is referenced by:  rabeqsn  4669  euabsn2  4729  dfiota2  6496  dfaiota2  45784  aiotaval  45793