Metamath Proof Explorer


Theorem prodeq12rdv

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

Ref Expression
Hypotheses prodeq12rdv.1
|- ( ph -> A = B )
prodeq12rdv.2
|- ( ( ph /\ k e. B ) -> C = D )
Assertion prodeq12rdv
|- ( ph -> prod_ k e. A C = prod_ k e. B D )

Proof

Step Hyp Ref Expression
1 prodeq12rdv.1
 |-  ( ph -> A = B )
2 prodeq12rdv.2
 |-  ( ( ph /\ k e. B ) -> C = D )
3 1 prodeq1d
 |-  ( ph -> prod_ k e. A C = prod_ k e. B C )
4 2 prodeq2dv
 |-  ( ph -> prod_ k e. B C = prod_ k e. B D )
5 3 4 eqtrd
 |-  ( ph -> prod_ k e. A C = prod_ k e. B D )