eldju2ndl - Metamath Proof Explorer

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Theorem eldju2ndl 9929

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 9906	. . . . 5 ⊢ (𝐴 ⊔ 𝐵) = (({∅} × 𝐴) ∪ ({1_o} × 𝐵))
2	1	eleq2i 2858	. . . 4 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ 𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)))
3		elun 4110	. . . 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 8029	. . . . 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 8029	. . . . 5 ⊢ (𝑋 ∈ ({1_o} × 𝐵) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)))
10		elsni 4611	. . . . . . 7 ⊢ ((1^st ‘𝑋) ∈ {1_o} → (1^st ‘𝑋) = 1_o)
11		1n0 8481	. . . . . . . 8 ⊢ 1_o ≠ ∅
12		neeq1 3023	. . . . . . . 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 2972	. . . . . . . 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 1570 ∈ wcel 2146 ≠ wne 2961 ∪ cun 3906 ∅c0 4289 {csn 4594 ⟨cop 4600 × cxp 5664 ‘cfv 6543 1^st c1st 7993 2^nd c2nd 7994 1_oc1o 8455 ⊔ cdju 9903

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 2148 ax-9 2156 ax-10 2179 ax-11 2195 ax-12 2216 ax-ext 2738 ax-sep 5262 ax-nul 5274 ax-pr 5409 ax-un 7745

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 2570 df-eu 2600 df-clab 2745 df-cleq 2758 df-clel 2841 df-nfc 2915 df-ne 2962 df-ral 3083 df-rex 3093 df-rab 3420 df-v 3460 df-dif 3911 df-un 3913 df-in 3915 df-ss 3925 df-nul 4290 df-if 4493 df-sn 4595 df-pr 4597 df-op 4601 df-uni 4878 df-br 5115 df-opab 5179 df-mpt 5198 df-id 5561 df-xp 5672 df-rel 5673 df-cnv 5674 df-co 5675 df-dm 5676 df-rn 5677 df-suc 6373 df-iota 6499 df-fun 6545 df-fv 6551 df-1st 7995 df-2nd 7996 df-1o 8462 df-dju 9906

This theorem is used by: updjudhf 9936

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