Theorem sbc5 1452

Description: An equivalence for class substitution.

Hypothesis

Ref	Expression
sbc5.1	⊢ A ∈ V

Assertion

Ref	Expression
sbc5	⊢ ([A / x]φ ↔ ∃x(x = A ∧ φ))

Distinct variable group(s):   x,A

Proof of Theorem sbc5

Step	Hyp	Ref	Expression
1		sbc5.1	. 2 ⊢ A ∈ V
2		sbc5g 1450	. 2 ⊢ (A ∈ V → ([A / x]φ ↔ ∃x(x = A ∧ φ)))
3	1, 2	ax-mp 6	1 ⊢ ([A / x]φ ↔ ∃x(x = A ∧ φ))

Syntax hints:   ↔ wb 127   ∧ wa 196  ∃wex 678   = wceq 1091   ∈ wcel 1092  Vcvv 1348  [wsbc 1440

This theorem is referenced by:  sbcie 1455

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-sbc 1441

metamath.org