Metamath Proof Explorer


Theorem prodeq2sdv

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

Ref Expression
Hypothesis prodeq2sdv.1
|- ( ph -> B = C )
Assertion prodeq2sdv
|- ( ph -> prod_ k e. A B = prod_ k e. A C )

Proof

Step Hyp Ref Expression
1 prodeq2sdv.1
 |-  ( ph -> B = C )
2 1 adantr
 |-  ( ( ph /\ k e. A ) -> B = C )
3 2 prodeq2dv
 |-  ( ph -> prod_ k e. A B = prod_ k e. A C )