Metamath Proof Explorer


Theorem prodeq12dv

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

Ref Expression
Hypotheses prodeq12dv.1 φ A = B
prodeq12dv.2 φ k A C = D
Assertion prodeq12dv φ k A C = k B D

Proof

Step Hyp Ref Expression
1 prodeq12dv.1 φ A = B
2 prodeq12dv.2 φ k A C = D
3 2 prodeq2dv φ k A C = k A D
4 1 prodeq1d φ k A D = k B D
5 3 4 eqtrd φ k A C = k B D