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Metamath Proof Explorer
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 ) ) |