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Theorem eldju2ndl 9919

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 9896	. . . . 5 ⊢ (𝐴 ⊔ 𝐵) = (({∅} × 𝐴) ∪ ({1_o} × 𝐵))
2	1	eleq2i 2826	. . . 4 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ 𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)))
3		elun 4149	. . . 4 ⊢ (𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)) ↔ (𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)))
4	2, 3	bitri 275	. . 3 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ (𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)))
5		elxp6 8009	. . . . 5 ⊢ (𝑋 ∈ ({∅} × 𝐴) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)))
6		simprr 772	. . . . . 6 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)) → (2^nd ‘𝑋) ∈ 𝐴)
7	6	a1d 25	. . . . 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 216	. . . 4 ⊢ (𝑋 ∈ ({∅} × 𝐴) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
9		elxp6 8009	. . . . 5 ⊢ (𝑋 ∈ ({1_o} × 𝐵) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)))
10		elsni 4646	. . . . . . 7 ⊢ ((1^st ‘𝑋) ∈ {1_o} → (1^st ‘𝑋) = 1_o)
11		1n0 8488	. . . . . . . 8 ⊢ 1_o ≠ ∅
12		neeq1 3004	. . . . . . . 8 ⊢ ((1^st ‘𝑋) = 1_o → ((1^st ‘𝑋) ≠ ∅ ↔ 1_o ≠ ∅))
13	11, 12	mpbiri 258	. . . . . . 7 ⊢ ((1^st ‘𝑋) = 1_o → (1^st ‘𝑋) ≠ ∅)
14		eqneqall 2952	. . . . . . . 8 ⊢ ((1^st ‘𝑋) = ∅ → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐴))
15	14	com12 32	. . . . . . 7 ⊢ ((1^st ‘𝑋) ≠ ∅ → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
16	10, 13, 15	3syl 18	. . . . . 6 ⊢ ((1^st ‘𝑋) ∈ {1_o} → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
17	16	ad2antrl 727	. . . . 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 216	. . . 4 ⊢ (𝑋 ∈ ({1_o} × 𝐵) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
19	8, 18	jaoi 856	. . 3 ⊢ ((𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
20	4, 19	sylbi 216	. 2 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
21	20	imp 408	1 ⊢ ((𝑋 ∈ (𝐴 ⊔ 𝐵) ∧ (1^st ‘𝑋) = ∅) → (2^nd ‘𝑋) ∈ 𝐴)

Colors of variables: wff setvar class

Syntax hints: → wi 4 ∧ wa 397 ∨ wo 846 = wceq 1542 ∈ wcel 2107 ≠ wne 2941 ∪ cun 3947 ∅c0 4323 {csn 4629 ⟨cop 4635 × cxp 5675 ‘cfv 6544 1^st c1st 7973 2^nd c2nd 7974 1_oc1o 8459 ⊔ cdju 9893

This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1914 ax-6 1972 ax-7 2012 ax-8 2109 ax-9 2117 ax-10 2138 ax-11 2155 ax-12 2172 ax-ext 2704 ax-sep 5300 ax-nul 5307 ax-pr 5428 ax-un 7725

This theorem depends on definitions: df-bi 206 df-an 398 df-or 847 df-3an 1090 df-tru 1545 df-fal 1555 df-ex 1783 df-nf 1787 df-sb 2069 df-mo 2535 df-eu 2564 df-clab 2711 df-cleq 2725 df-clel 2811 df-nfc 2886 df-ne 2942 df-ral 3063 df-rex 3072 df-rab 3434 df-v 3477 df-dif 3952 df-un 3954 df-in 3956 df-ss 3966 df-nul 4324 df-if 4530 df-sn 4630 df-pr 4632 df-op 4636 df-uni 4910 df-br 5150 df-opab 5212 df-mpt 5233 df-id 5575 df-xp 5683 df-rel 5684 df-cnv 5685 df-co 5686 df-dm 5687 df-rn 5688 df-suc 6371 df-iota 6496 df-fun 6546 df-fv 6552 df-1st 7975 df-2nd 7976 df-1o 8466 df-dju 9896

This theorem is referenced by: updjudhf 9926

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