Theorem sbc5 3775

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 3748	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
2		clelab 2910	. 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 2146  {cab 2744  [wsbc 3747

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-12 2216  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-sbc 3748

This theorem is used by:  sbc7  3779  sbccomlemOLD  3826  csb2  3858  rexsns  4642  sbcop1  5475  sbccom2lem  38814  pm13.192  45161  pm13.195  45164  2sbc5g  45167  iotasbc  45170  pm14.122b  45174  iotasbc5  45182  sbcpr  48311