eldju2ndr - Metamath Proof Explorer

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Theorem eldju2ndr 9931

Description: The second component of an element of a disjoint union is an element of the right class of the disjoint union if its first component is not the empty set. (Contributed by AV, 26-Jun-2022.)

**Assertion**
Ref	Expression
eldju2ndr	⊢ ((𝑋 ∈ (𝐴 ⊔ 𝐵) ∧ (1^st ‘𝑋) ≠ ∅) → (2^nd ‘𝑋) ∈ 𝐵)

**Proof of Theorem eldju2ndr**
Step	Hyp	Ref	Expression
1		df-dju 9907	. . . . 5 ⊢ (𝐴 ⊔ 𝐵) = (({∅} × 𝐴) ∪ ({1_o} × 𝐵))
2	1	eleq2i 2852	. . . 4 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ 𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)))
3		elun 4100	. . . 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 8021	. . . . 5 ⊢ (𝑋 ∈ ({∅} × 𝐴) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)))
6		elsni 4601	. . . . . . 7 ⊢ ((1^st ‘𝑋) ∈ {∅} → (1^st ‘𝑋) = ∅)
7		eqneqall 2966	. . . . . . 7 ⊢ ((1^st ‘𝑋) = ∅ → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
8	6, 7	syl 18	. . . . . 6 ⊢ ((1^st ‘𝑋) ∈ {∅} → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
9	8	ad2antrl 741	. . . . 5 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)) → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
10	5, 9	sylbi 220	. . . 4 ⊢ (𝑋 ∈ ({∅} × 𝐴) → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
11		elxp6 8021	. . . . 5 ⊢ (𝑋 ∈ ({1_o} × 𝐵) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)))
12		simprr 785	. . . . . 6 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)) → (2^nd ‘𝑋) ∈ 𝐵)
13	12	a1d 26	. . . . 5 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)) → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
14	11, 13	sylbi 220	. . . 4 ⊢ (𝑋 ∈ ({1_o} × 𝐵) → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
15	10, 14	jaoi 871	. . 3 ⊢ ((𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)) → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
16	4, 15	sylbi 220	. 2 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐵))
17	16	imp 412	1 ⊢ ((𝑋 ∈ (𝐴 ⊔ 𝐵) ∧ (1^st ‘𝑋) ≠ ∅) → (2^nd ‘𝑋) ∈ 𝐵)

Colors of variables: wff setvar class

This proof depends on syntax axioms: → wi 4 ∧ wa 401 ∨ wo 861 = wceq 1570 ∈ wcel 2145 ≠ wne 2955 ∪ cun 3897 ∅c0 4279 {csn 4584 ⟨cop 4590 × cxp 5653 ‘cfv 6533 1^st c1st 7985 2^nd c2nd 7986 1_oc1o 8449 ⊔ cdju 9904

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-10 2178 ax-11 2194 ax-12 2213 ax-ext 2732 ax-sep 5251 ax-nul 5263 ax-pr 5398 ax-un 7737

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-nf 1817 df-sb 2100 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-dif 3902 df-un 3904 df-in 3906 df-ss 3916 df-nul 4280 df-if 4483 df-sn 4585 df-pr 4587 df-op 4591 df-uni 4868 df-br 5104 df-opab 5168 df-mpt 5187 df-id 5550 df-xp 5661 df-rel 5662 df-cnv 5663 df-co 5664 df-dm 5665 df-rn 5666 df-iota 6489 df-fun 6535 df-fv 6541 df-1st 7987 df-2nd 7988 df-dju 9907

This theorem is used by: updjudhf 9937

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