Theorem sqxpeqd 4690

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 4689	1 ⊢ (𝜑 → (𝐴 × 𝐴) = (𝐵 × 𝐵))

Syntax hints:   → wi 4   = wceq 1364   × cxp 4662

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 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-11 1520  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-opab 4096  df-xp 4670

This theorem is referenced by:  prdsval  12975  imasaddfnlemg  13016  intopsn  13069  srg1zr  13619  ispsmet  14643  isxms  14771  isms  14773  xmspropd  14797  mspropd  14798