Theorem sbc5 3793

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 3766	. 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 1540  ∃wex 1779   ∈ wcel 2108  {cab 2713  [wsbc 3765

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-12 2177  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-sbc 3766

This theorem is referenced by:  sbc7  3797  sbciegftOLD  3803  sbccomlemOLD  3845  csb2  3876  rexsns  4647  sbcop1  5463  sbccom2lem  38148  pm13.192  44434  pm13.195  44437  2sbc5g  44440  iotasbc  44443  pm14.122b  44447  iotasbc5  44455  sbcpr  47535