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 φ A = B
Assertion sqxpeqd φ A × A = B × B

Proof

Step Hyp Ref Expression
1 xpeq1d.1 φ A = B
2 1 1 xpeq12d φ A × A = B × B