Theorem sbc5 3756

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 3729	. 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 1542  ∃wex 1781   ∈ wcel 2114  {cab 2714  [wsbc 3728

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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-12 2185  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-sbc 3729

This theorem is referenced by:  sbc7  3760  sbciegftOLD  3766  sbccomlemOLD  3808  csb2  3839  rexsns  4615  sbcop1  5441  sbccom2lem  38445  pm13.192  44837  pm13.195  44840  2sbc5g  44843  iotasbc  44846  pm14.122b  44850  iotasbc5  44858  sbcpr  47981