Theorem fnxpdmdm 48651

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 6596	. 2 ⊢ (𝐹 Fn (𝐴 × 𝐴) → dom 𝐹 = (𝐴 × 𝐴))
2		dmeq 5853	. . 3 ⊢ (dom 𝐹 = (𝐴 × 𝐴) → dom dom 𝐹 = dom (𝐴 × 𝐴))
3		dmxpid 5880	. . 3 ⊢ dom (𝐴 × 𝐴) = 𝐴
4	2, 3	eqtrdi 2788	. 2 ⊢ (dom 𝐹 = (𝐴 × 𝐴) → dom dom 𝐹 = 𝐴)
5	1, 4	syl 17	1 ⊢ (𝐹 Fn (𝐴 × 𝐴) → dom dom 𝐹 = 𝐴)

Syntax hints:   → wi 4   = wceq 1542   × cxp 5623  dom cdm 5625   Fn wfn 6488

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5232  ax-pr 5371

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-ral 3053  df-rex 3063  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-br 5087  df-opab 5149  df-xp 5631  df-dm 5635  df-fn 6496

This theorem is referenced by: (None)