Metamath Proof Explorer


Theorem prodeq12rdv

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

Ref Expression
Hypotheses prodeq12rdv.1 φ A = B
prodeq12rdv.2 φ k B C = D
Assertion prodeq12rdv φ k A C = k B D

Proof

Step Hyp Ref Expression
1 prodeq12rdv.1 φ A = B
2 prodeq12rdv.2 φ k B C = D
3 1 prodeq1d φ k A C = k B C
4 2 prodeq2dv φ k B C = k B D
5 3 4 eqtrd φ k A C = k B D