Theorem sbc5 3770

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 3743	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
2		clelab 2908	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))
3	1, 2	bitri 278	1 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2147  {cab 2741  [wsbc 3742

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-sbc 3743

This theorem is used by:  sbc7  3774  csb2  3853  rexsns  4632  sbcop1  5463  sbccom2lem  39259  pm13.192  45597  pm13.195  45600  2sbc5g  45603  iotasbc  45605  pm14.122b  45609  iotasbc5  45617  sbcpr  48818