Theorem sqxpeqd 4800

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 4799	1 ⊢ (𝜑 → (𝐴 × 𝐴) = (𝐵 × 𝐵))

This proof depends on syntax axioms:   → wi 4   = wceq 1402   × cxp 4772

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-11 1559  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-opab 4193  df-xp 4780

This theorem is used by:  papeq2  7610  imasaddfnlemg  13635  intopsn  13687  prdsval  14173  rng1zrlem  14258  ispsmet  15424  isxms  15552  isms  15554  xmspropd  15578  mspropd  15579