Theorem sbc5 3768

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 3741	. 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 206   ∧ wa 395   = wceq 1541  ∃wex 1780   ∈ wcel 2113  {cab 2714  [wsbc 3740

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-10 2146  ax-12 2184  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2715  df-cleq 2728  df-clel 2811  df-sbc 3741

This theorem is referenced by:  sbc7  3772  sbciegftOLD  3778  sbccomlemOLD  3820  csb2  3851  rexsns  4628  sbcop1  5436  sbccom2lem  38321  pm13.192  44647  pm13.195  44650  2sbc5g  44653  iotasbc  44656  pm14.122b  44660  iotasbc5  44668  sbcpr  47763