Metamath Proof Explorer


Theorem prodeq2dv

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

Ref Expression
Hypothesis prodeq2dv.1
|- ( ( ph /\ k e. A ) -> B = C )
Assertion prodeq2dv
|- ( ph -> prod_ k e. A B = prod_ k e. A C )

Proof

Step Hyp Ref Expression
1 prodeq2dv.1
 |-  ( ( ph /\ k e. A ) -> B = C )
2 1 ralrimiva
 |-  ( ph -> A. k e. A B = C )
3 2 prodeq2d
 |-  ( ph -> prod_ k e. A B = prod_ k e. A C )