Theorem iunmapdisj 8791

Description: The union ∪ 𝑛 ∈ 𝐶(𝐴 ↑_𝑚 𝑛) is a disjoint union. (Contributed by Mario Carneiro, 17-May-2015.) (Revised by NM, 16-Jun-2017.)

Assertion

Ref	Expression
iunmapdisj	⊢ ∃*𝑛 ∈ 𝐶 𝐵 ∈ (𝐴 ↑𝑚 𝑛)

Distinct variable group:   𝐵,𝑛

Allowed substitution hints:   𝐴(𝑛)   𝐶(𝑛)

Proof of Theorem iunmapdisj

Step	Hyp	Ref	Expression
1		moeq 3369	. . . 4 ⊢ ∃*𝑛 𝑛 = dom 𝐵
2		elmapi 7824	. . . . . 6 ⊢ (𝐵 ∈ (𝐴 ↑𝑚 𝑛) → 𝐵:𝑛⟶𝐴)
3		fdm 6010	. . . . . . 7 ⊢ (𝐵:𝑛⟶𝐴 → dom 𝐵 = 𝑛)
4	3	eqcomd 2632	. . . . . 6 ⊢ (𝐵:𝑛⟶𝐴 → 𝑛 = dom 𝐵)
5	2, 4	syl 17	. . . . 5 ⊢ (𝐵 ∈ (𝐴 ↑𝑚 𝑛) → 𝑛 = dom 𝐵)
6	5	moimi 2524	. . . 4 ⊢ (∃*𝑛 𝑛 = dom 𝐵 → ∃*𝑛 𝐵 ∈ (𝐴 ↑𝑚 𝑛))
7	1, 6	ax-mp 5	. . 3 ⊢ ∃*𝑛 𝐵 ∈ (𝐴 ↑𝑚 𝑛)
8	7	moani 2529	. 2 ⊢ ∃*𝑛(𝑛 ∈ 𝐶 ∧ 𝐵 ∈ (𝐴 ↑𝑚 𝑛))
9		df-rmo 2920	. 2 ⊢ (∃*𝑛 ∈ 𝐶 𝐵 ∈ (𝐴 ↑𝑚 𝑛) ↔ ∃*𝑛(𝑛 ∈ 𝐶 ∧ 𝐵 ∈ (𝐴 ↑𝑚 𝑛)))
10	8, 9	mpbir 221	1 ⊢ ∃*𝑛 ∈ 𝐶 𝐵 ∈ (𝐴 ↑𝑚 𝑛)

Syntax hints:   ∧ wa 384   = wceq 1480   ∈ wcel 1992  ∃*wmo 2475  ∃*wrmo 2915  dom cdm 5079  ⟶wf 5846  (class class class)co 6605   ↑_𝑚 cmap 7803

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1841  ax-6 1890  ax-7 1937  ax-8 1994  ax-9 2001  ax-10 2021  ax-11 2036  ax-12 2049  ax-13 2250  ax-ext 2606  ax-sep 4746  ax-nul 4754  ax-pow 4808  ax-pr 4872  ax-un 6903

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1038  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1883  df-eu 2478  df-mo 2479  df-clab 2613  df-cleq 2619  df-clel 2622  df-nfc 2756  df-ne 2797  df-ral 2917  df-rex 2918  df-rmo 2920  df-rab 2921  df-v 3193  df-sbc 3423  df-csb 3520  df-dif 3563  df-un 3565  df-in 3567  df-ss 3574  df-nul 3897  df-if 4064  df-pw 4137  df-sn 4154  df-pr 4156  df-op 4160  df-uni 4408  df-iun 4492  df-br 4619  df-opab 4679  df-mpt 4680  df-id 4994  df-xp 5085  df-rel 5086  df-cnv 5087  df-co 5088  df-dm 5089  df-rn 5090  df-res 5091  df-ima 5092  df-iota 5813  df-fun 5852  df-fn 5853  df-f 5854  df-fv 5858  df-ov 6608  df-oprab 6609  df-mpt2 6610  df-1st 7116  df-2nd 7117  df-map 7805

This theorem is referenced by: (None)