Metamath Proof Explorer


Theorem caovdirg

Description: Convert an operation reverse distributive law to class notation. (Contributed by Mario Carneiro, 19-Oct-2014)

Ref Expression
Hypothesis caovdirg.1
|- ( ( ph /\ ( x e. S /\ y e. S /\ z e. K ) ) -> ( ( x F y ) G z ) = ( ( x G z ) H ( y G z ) ) )
Assertion caovdirg
|- ( ( ph /\ ( A e. S /\ B e. S /\ C e. K ) ) -> ( ( A F B ) G C ) = ( ( A G C ) H ( B G C ) ) )

Proof

Step Hyp Ref Expression
1 caovdirg.1
 |-  ( ( ph /\ ( x e. S /\ y e. S /\ z e. K ) ) -> ( ( x F y ) G z ) = ( ( x G z ) H ( y G z ) ) )
2 1 ralrimivvva
 |-  ( ph -> A. x e. S A. y e. S A. z e. K ( ( x F y ) G z ) = ( ( x G z ) H ( y G z ) ) )
3 oveq1
 |-  ( x = A -> ( x F y ) = ( A F y ) )
4 3 oveq1d
 |-  ( x = A -> ( ( x F y ) G z ) = ( ( A F y ) G z ) )
5 oveq1
 |-  ( x = A -> ( x G z ) = ( A G z ) )
6 5 oveq1d
 |-  ( x = A -> ( ( x G z ) H ( y G z ) ) = ( ( A G z ) H ( y G z ) ) )
7 4 6 eqeq12d
 |-  ( x = A -> ( ( ( x F y ) G z ) = ( ( x G z ) H ( y G z ) ) <-> ( ( A F y ) G z ) = ( ( A G z ) H ( y G z ) ) ) )
8 oveq2
 |-  ( y = B -> ( A F y ) = ( A F B ) )
9 8 oveq1d
 |-  ( y = B -> ( ( A F y ) G z ) = ( ( A F B ) G z ) )
10 oveq1
 |-  ( y = B -> ( y G z ) = ( B G z ) )
11 10 oveq2d
 |-  ( y = B -> ( ( A G z ) H ( y G z ) ) = ( ( A G z ) H ( B G z ) ) )
12 9 11 eqeq12d
 |-  ( y = B -> ( ( ( A F y ) G z ) = ( ( A G z ) H ( y G z ) ) <-> ( ( A F B ) G z ) = ( ( A G z ) H ( B G z ) ) ) )
13 oveq2
 |-  ( z = C -> ( ( A F B ) G z ) = ( ( A F B ) G C ) )
14 oveq2
 |-  ( z = C -> ( A G z ) = ( A G C ) )
15 oveq2
 |-  ( z = C -> ( B G z ) = ( B G C ) )
16 14 15 oveq12d
 |-  ( z = C -> ( ( A G z ) H ( B G z ) ) = ( ( A G C ) H ( B G C ) ) )
17 13 16 eqeq12d
 |-  ( z = C -> ( ( ( A F B ) G z ) = ( ( A G z ) H ( B G z ) ) <-> ( ( A F B ) G C ) = ( ( A G C ) H ( B G C ) ) ) )
18 7 12 17 rspc3v
 |-  ( ( A e. S /\ B e. S /\ C e. K ) -> ( A. x e. S A. y e. S A. z e. K ( ( x F y ) G z ) = ( ( x G z ) H ( y G z ) ) -> ( ( A F B ) G C ) = ( ( A G C ) H ( B G C ) ) ) )
19 2 18 mpan9
 |-  ( ( ph /\ ( A e. S /\ B e. S /\ C e. K ) ) -> ( ( A F B ) G C ) = ( ( A G C ) H ( B G C ) ) )