Theorem sbc5 3758

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 3731	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
2		clelab 2884	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))
3	1, 2	bitri 276	1 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ↔ wb 207   ∧ wa 396   = wceq 1547  ∃wex 1786   ∈ wcel 2119  {cab 2718  [wsbc 3730

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-10 2152  ax-12 2189  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-sbc 3731

This theorem is referenced by:  sbc7  3762  sbccomlemOLD  3809  csb2  3840  rexsns  4610  sbcop1  5435  sbccom2lem  38498  pm13.192  44861  pm13.195  44864  2sbc5g  44867  iotasbc  44870  pm14.122b  44874  iotasbc5  44882  sbcpr  48003