Theorem absn 4611

Description: Condition for a class abstraction to be a singleton. Formerly part of proof of dfiota2 6497. (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 4592	. . 3 ⊢ {𝑌} = {𝑥 ∣ 𝑥 = 𝑌}
2	1	eqeq2i 2778	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌})
3		abbib 2834	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))
4	2, 3	bitri 278	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568   = wceq 1570  {cab 2743  {csn 4591

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-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-sn 4592

This theorem is used by:  rabeqsn  4635  euabsn2  4693  dfiota2  6497  n0cut  28582  dfaiota2  47900  aiotaval  47909