Metamath Proof Explorer


Theorem prodeq12dv

Description: Equality deduction for product. (Contributed by Scott Fenton, 4-Dec-2017)

Ref Expression
Hypotheses prodeq12dv.1
|- ( ph -> A = B )
prodeq12dv.2
|- ( ( ph /\ k e. A ) -> C = D )
Assertion prodeq12dv
|- ( ph -> prod_ k e. A C = prod_ k e. B D )

Proof

Step Hyp Ref Expression
1 prodeq12dv.1
 |-  ( ph -> A = B )
2 prodeq12dv.2
 |-  ( ( ph /\ k e. A ) -> C = D )
3 2 prodeq2dv
 |-  ( ph -> prod_ k e. A C = prod_ k e. A D )
4 1 prodeq1d
 |-  ( ph -> prod_ k e. A D = prod_ k e. B D )
5 3 4 eqtrd
 |-  ( ph -> prod_ k e. A C = prod_ k e. B D )