eldju2ndl - Metamath Proof Explorer

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Theorem eldju2ndl 10005

Description: The second component of an element of a disjoint union is an element of the left class of the disjoint union if its first component is the empty set. (Contributed by AV, 26-Jun-2022.)

**Assertion**
Ref	Expression
eldju2ndl	⊢ ((𝑋 ∈ (𝐴 ⊔ 𝐵) ∧ (1^st ‘𝑋) = ∅) → (2^nd ‘𝑋) ∈ 𝐴)

**Proof of Theorem eldju2ndl**
Step	Hyp	Ref	Expression
1		df-dju 9982	. . . . 5 ⊢ (𝐴 ⊔ 𝐵) = (({∅} × 𝐴) ∪ ({1_o} × 𝐵))
2	1	eleq2i 2852	. . . 4 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ 𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)))
3		elun 4152	. . . 4 ⊢ (𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)) ↔ (𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)))
4	2, 3	bitri 278	. . 3 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ (𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)))
5		elxp6 8035	. . . . 5 ⊢ (𝑋 ∈ ({∅} × 𝐴) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)))
6		simprr 785	. . . . . 6 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)) → (2^nd ‘𝑋) ∈ 𝐴)
7	6	a1d 26	. . . . 5 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
8	5, 7	sylbi 220	. . . 4 ⊢ (𝑋 ∈ ({∅} × 𝐴) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
9		elxp6 8035	. . . . 5 ⊢ (𝑋 ∈ ({1_o} × 𝐵) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)))
10		elsni 4600	. . . . . . 7 ⊢ ((1^st ‘𝑋) ∈ {1_o} → (1^st ‘𝑋) = 1_o)
11		1n0 8495	. . . . . . . 8 ⊢ 1_o ≠ ∅
12		neeq1 3017	. . . . . . . 8 ⊢ ((1^st ‘𝑋) = 1_o → ((1^st ‘𝑋) ≠ ∅ ↔ 1_o ≠ ∅))
13	11, 12	mpbiri 261	. . . . . . 7 ⊢ ((1^st ‘𝑋) = 1_o → (1^st ‘𝑋) ≠ ∅)
14		eqneqall 2966	. . . . . . . 8 ⊢ ((1^st ‘𝑋) = ∅ → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐴))
15	14	com12 33	. . . . . . 7 ⊢ ((1^st ‘𝑋) ≠ ∅ → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
16	10, 13, 15	3syl 19	. . . . . 6 ⊢ ((1^st ‘𝑋) ∈ {1_o} → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
17	16	ad2antrl 741	. . . . 5 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
18	9, 17	sylbi 220	. . . 4 ⊢ (𝑋 ∈ ({1_o} × 𝐵) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
19	8, 18	jaoi 871	. . 3 ⊢ ((𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
20	4, 19	sylbi 220	. 2 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
21	20	imp 412	1 ⊢ ((𝑋 ∈ (𝐴 ⊔ 𝐵) ∧ (1^st ‘𝑋) = ∅) → (2^nd ‘𝑋) ∈ 𝐴)

Colors of variables: wff setvar class

This proof depends on syntax axioms: → wi 4 ∧ wa 401 ∨ wo 861 = wceq 1569 ∈ wcel 2144 ≠ wne 2955 ∪ cun 4149 ∅c0 4278 {csn 4583 ⟨cop 4589 × cxp 5648 ‘cfv 6538 1^st c1st 7999 2^nd c2nd 8000 1_oc1o 8469 ⊔ cdju 9979

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 1942 ax-6 1999 ax-7 2040 ax-8 2146 ax-9 2154 ax-10 2177 ax-11 2193 ax-12 2212 ax-ext 2732 ax-sep 5248 ax-nul 5259 ax-pr 5390 ax-un 7751

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-nf 1816 df-sb 2099 df-mo 2564 df-eu 2594 df-clab 2739 df-cleq 2752 df-clel 2835 df-nfc 2909 df-ne 2956 df-ral 3077 df-rex 3087 df-rab 3413 df-v 3452 df-ss 3896 df-dif 4047 df-in 4091 df-un 4151 df-nul 4279 df-if 4482 df-sn 4584 df-pr 4586 df-op 4590 df-uni 4867 df-br 5103 df-opab 5167 df-mpt 5186 df-id 5545 df-xp 5656 df-rel 5657 df-cnv 5658 df-co 5659 df-dm 5660 df-rn 5661 df-suc 6368 df-iota 6494 df-fun 6540 df-fv 6546 df-1st 8001 df-2nd 8002 df-1o 8476 df-dju 9982

This theorem is used by: updjudhf 10012

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