Theorem dmdm 49987

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

Step	Hyp	Ref	Expression
1		fndm 6639	. . 3 ⊢ (𝐴 Fn (𝐵 × 𝐵) → dom 𝐴 = (𝐵 × 𝐵))
2	1	dmeqd 5893	. 2 ⊢ (𝐴 Fn (𝐵 × 𝐵) → dom dom 𝐴 = dom (𝐵 × 𝐵))
3		dmxpid 5918	. 2 ⊢ dom (𝐵 × 𝐵) = 𝐵
4	2, 3	eqtr2di 2814	1 ⊢ (𝐴 Fn (𝐵 × 𝐵) → 𝐵 = dom dom 𝐴)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   × cxp 5657  dom cdm 5659   Fn wfn 6532

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-sep 5255  ax-pr 5402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-br 5108  df-opab 5172  df-xp 5665  df-dm 5669  df-fn 6540

This theorem is used by:  iinfconstbas  50000