Theorem absn 4642

Description: Condition for a class abstraction to be a singleton. Formerly part of proof of dfiota2 6495. (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 4625	. . 3 ⊢ {𝑌} = {𝑥 ∣ 𝑥 = 𝑌}
2	1	eqeq2i 2741	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌})
3		abbib 2800	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))
4	2, 3	bitri 275	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))

Syntax hints:   ↔ wb 205  ∀wal 1532   = wceq 1534  {cab 2705  {csn 4624

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2706  df-cleq 2720  df-sn 4625

This theorem is referenced by:  rabeqsn  4665  euabsn2  4725  dfiota2  6495  n0scut  28196  dfaiota2  46460  aiotaval  46469