Theorem sbc5 3806

Description: An equivalence for class substitution. (Contributed by NM, 23-Aug-1993.) (Revised by Mario Carneiro, 12-Oct-2016.) (Proof shortened by SN, 2-Sep-2024.)

Assertion

Ref	Expression
sbc5	⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem sbc5

Step	Hyp	Ref	Expression
1		df-sbc 3779	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
2		clelab 2880	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))
3	1, 2	bitri 275	1 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ↔ wb 205   ∧ wa 397   = wceq 1542  ∃wex 1782   ∈ wcel 2107  {cab 2710  [wsbc 3778

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-12 2172  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-sbc 3779

This theorem is referenced by:  sbc6gOLD  3809  sbc7  3811  sbciegft  3816  sbccomlem  3865  csb2  3896  rexsns  4674  sbcop1  5489  sbccom2lem  36992  pm13.192  43169  pm13.195  43172  2sbc5g  43175  iotasbc  43178  pm14.122b  43182  iotasbc5  43190  sbcpr  46189