Metamath Proof Explorer


Theorem csb2

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)

Ref Expression
Assertion csb2
|- [_ A / x ]_ B = { y | E. x ( x = A /\ y e. B ) }

Proof

Step Hyp Ref Expression
1 df-csb
 |-  [_ A / x ]_ B = { y | [. A / x ]. y e. B }
2 sbc5
 |-  ( [. A / x ]. y e. B <-> E. x ( x = A /\ y e. B ) )
3 2 abbii
 |-  { y | [. A / x ]. y e. B } = { y | E. x ( x = A /\ y e. B ) }
4 1 3 eqtri
 |-  [_ A / x ]_ B = { y | E. x ( x = A /\ y e. B ) }