Metamath Proof Explorer


Theorem sqxpeqd

Description: Equality deduction for a Cartesian square, see Wikipedia "Cartesian product", https://en.wikipedia.org/wiki/Cartesian_product#n-ary_Cartesian_power . (Contributed by AV, 13-Jan-2020)

Ref Expression
Hypothesis xpeq1d.1
|- ( ph -> A = B )
Assertion sqxpeqd
|- ( ph -> ( A X. A ) = ( B X. B ) )

Proof

Step Hyp Ref Expression
1 xpeq1d.1
 |-  ( ph -> A = B )
2 1 1 xpeq12d
 |-  ( ph -> ( A X. A ) = ( B X. B ) )