Theorem sqxpeqd 4664

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.)

Hypothesis

Ref	Expression
xpeq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
sqxpeqd	⊢ (𝜑 → (𝐴 × 𝐴) = (𝐵 × 𝐵))

Proof of Theorem sqxpeqd

Step	Hyp	Ref	Expression
1		xpeq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2	1, 1	xpeq12d 4663	1 ⊢ (𝜑 → (𝐴 × 𝐴) = (𝐵 × 𝐵))

Syntax hints:   → wi 4   = wceq 1363   × cxp 4636

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-11 1516  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169

This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-opab 4077  df-xp 4644

This theorem is referenced by:  imasaddfnlemg  12752  intopsn  12804  srg1zr  13234  ispsmet  14094  isxms  14222  isms  14224  xmspropd  14248  mspropd  14249