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Theorem dmdm 49035
Description: The double domain of a function on a Cartesian square. (Contributed by Zhi Wang, 1-Nov-2025.)
Assertion
Ref Expression
dmdm (𝐴 Fn (𝐵 × 𝐵) → 𝐵 = dom dom 𝐴)

Proof of Theorem dmdm
StepHypRef Expression
1 fndm 6603 . . 3 (𝐴 Fn (𝐵 × 𝐵) → dom 𝐴 = (𝐵 × 𝐵))
21dmeqd 5859 . 2 (𝐴 Fn (𝐵 × 𝐵) → dom dom 𝐴 = dom (𝐵 × 𝐵))
3 dmxpid 5883 . 2 dom (𝐵 × 𝐵) = 𝐵
42, 3eqtr2di 2781 1 (𝐴 Fn (𝐵 × 𝐵) → 𝐵 = dom dom 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1540   × cxp 5629  dom cdm 5631   Fn wfn 6494
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701  ax-sep 5246  ax-nul 5256  ax-pr 5382
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-ne 2926  df-ral 3045  df-rex 3054  df-rab 3403  df-v 3446  df-dif 3914  df-un 3916  df-ss 3928  df-nul 4293  df-if 4485  df-sn 4586  df-pr 4588  df-op 4592  df-br 5103  df-opab 5165  df-xp 5637  df-dm 5641  df-fn 6502
This theorem is referenced by:  iinfconstbas  49048
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