Theorem csb2 2910

Description: Alternate expression for the proper substitution into a class, without referencing substitution into a wff. Note that 𝑥 can be free in 𝐵 but cannot occur in 𝐴. (Contributed by NM, 2-Dec-2013.)

Assertion

Ref	Expression
csb2	⊢ ⦋𝐴 / 𝑥⦌𝐵 = {𝑦 ∣ ∃𝑥(𝑥 = 𝐴 ∧ 𝑦 ∈ 𝐵)}

Distinct variable groups:   𝑥,𝑦,𝐴   𝑦,𝐵

Allowed substitution hint:   𝐵(𝑥)

Proof of Theorem csb2

Step	Hyp	Ref	Expression
1		df-csb 2909	. 2 ⊢ ⦋𝐴 / 𝑥⦌𝐵 = {𝑦 ∣ [𝐴 / 𝑥]𝑦 ∈ 𝐵}
2		sbc5 2838	. . 3 ⊢ ([𝐴 / 𝑥]𝑦 ∈ 𝐵 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝑦 ∈ 𝐵))
3	2	abbii 2194	. 2 ⊢ {𝑦 ∣ [𝐴 / 𝑥]𝑦 ∈ 𝐵} = {𝑦 ∣ ∃𝑥(𝑥 = 𝐴 ∧ 𝑦 ∈ 𝐵)}
4	1, 3	eqtri 2101	1 ⊢ ⦋𝐴 / 𝑥⦌𝐵 = {𝑦 ∣ ∃𝑥(𝑥 = 𝐴 ∧ 𝑦 ∈ 𝐵)}

Syntax hints:   ∧ wa 102   = wceq 1284  ∃wex 1421   ∈ wcel 1433  {cab 2067  [wsbc 2815  ⦋csb 2908

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-sbc 2816  df-csb 2909

This theorem is referenced by: (None)