Theorem sbc5 3769

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 3742	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
2		clelab 2876	. 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 1541  ∃wex 1780   ∈ wcel 2111  {cab 2709  [wsbc 3741

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-12 2180  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-sbc 3742

This theorem is referenced by:  sbc7  3773  sbciegftOLD  3779  sbccomlemOLD  3821  csb2  3852  rexsns  4624  sbcop1  5428  sbccom2lem  38163  pm13.192  44442  pm13.195  44445  2sbc5g  44448  iotasbc  44451  pm14.122b  44455  iotasbc5  44463  sbcpr  47551