Theorem fnxpdmdm 49402

Description: The domain of the domain of a function over a Cartesian square. (Contributed by AV, 13-Jan-2020.)

Assertion

Ref	Expression
fnxpdmdm	⊢ (𝐹 Fn (𝐴 × 𝐴) → dom dom 𝐹 = 𝐴)

Proof of Theorem fnxpdmdm

Step	Hyp	Ref	Expression
1		fndm 6653	. 2 ⊢ (𝐹 Fn (𝐴 × 𝐴) → dom 𝐹 = (𝐴 × 𝐴))
2		dmeq 5890	. . 3 ⊢ (dom 𝐹 = (𝐴 × 𝐴) → dom dom 𝐹 = dom (𝐴 × 𝐴))
3		dmxpid 5917	. . 3 ⊢ dom (𝐴 × 𝐴) = 𝐴
4	2, 3	eqtrdi 2814	. 2 ⊢ (dom 𝐹 = (𝐴 × 𝐴) → dom dom 𝐹 = 𝐴)
5	1, 4	syl 18	1 ⊢ (𝐹 Fn (𝐴 × 𝐴) → dom dom 𝐹 = 𝐴)

This proof depends on syntax axioms:   → wi 4   = wceq 1569   × cxp 5653  dom cdm 5655   Fn wfn 6542

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735  ax-sep 5252  ax-pr 5395

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5171  df-xp 5661  df-dm 5665  df-fn 6550

This theorem is used by: (None)