Theorem sqxpeqd 4702

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	|- ( ph -> A = B )

Assertion

Ref	Expression
sqxpeqd	|- ( ph -> ( A X. A ) = ( B X. B ) )

Proof of Theorem sqxpeqd

Step	Hyp	Ref	Expression
1		xpeq1d.1	. 2 |- ( ph -> A = B )
2	1, 1	xpeq12d 4701	1 |- ( ph -> ( A X. A ) = ( B X. B ) )

Syntax hints:   -> wi 4   = wceq 1373   X. cxp 4674

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 4107  df-xp 4682

This theorem is referenced by:  prdsval  13138  imasaddfnlemg  13179  intopsn  13232  srg1zr  13782  ispsmet  14828  isxms  14956  isms  14958  xmspropd  14982  mspropd  14983