Theorem sqxpeqd 4701

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 4700	1 ⊢ (𝜑 → (𝐴 × 𝐴) = (𝐵 × 𝐵))

Syntax hints:   → wi 4   = wceq 1373   × cxp 4673

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 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-11 1529  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201  df-opab 4106  df-xp 4681

This theorem is referenced by:  prdsval  13105  imasaddfnlemg  13146  intopsn  13199  srg1zr  13749  ispsmet  14795  isxms  14923  isms  14925  xmspropd  14949  mspropd  14950