Metamath Proof Explorer


Theorem csbco

Description: Composition law for chained substitutions into a class. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker csbcow when possible. (Contributed by NM, 10-Nov-2005) (New usage is discouraged.)

Ref Expression
Assertion csbco
|- [_ A / y ]_ [_ y / x ]_ B = [_ A / x ]_ B

Proof

Step Hyp Ref Expression
1 df-csb
 |-  [_ y / x ]_ B = { z | [. y / x ]. z e. B }
2 1 abeq2i
 |-  ( z e. [_ y / x ]_ B <-> [. y / x ]. z e. B )
3 2 sbcbii
 |-  ( [. A / y ]. z e. [_ y / x ]_ B <-> [. A / y ]. [. y / x ]. z e. B )
4 sbcco
 |-  ( [. A / y ]. [. y / x ]. z e. B <-> [. A / x ]. z e. B )
5 3 4 bitri
 |-  ( [. A / y ]. z e. [_ y / x ]_ B <-> [. A / x ]. z e. B )
6 5 abbii
 |-  { z | [. A / y ]. z e. [_ y / x ]_ B } = { z | [. A / x ]. z e. B }
7 df-csb
 |-  [_ A / y ]_ [_ y / x ]_ B = { z | [. A / y ]. z e. [_ y / x ]_ B }
8 df-csb
 |-  [_ A / x ]_ B = { z | [. A / x ]. z e. B }
9 6 7 8 3eqtr4i
 |-  [_ A / y ]_ [_ y / x ]_ B = [_ A / x ]_ B