eldju2ndl - Intuitionistic Logic Explorer

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Theorem eldju2ndl 7412

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 7378	. . . . 5 ⊢ (𝐴 ⊔ 𝐵) = (({∅} × 𝐴) ∪ ({1_o} × 𝐵))
2	1	eleq2i 2305	. . . 4 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ 𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)))
3		elun 3370	. . . 4 ⊢ (𝑋 ∈ (({∅} × 𝐴) ∪ ({1_o} × 𝐵)) ↔ (𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)))
4	2, 3	bitri 184	. . 3 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) ↔ (𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)))
5		elxp6 6403	. . . . 5 ⊢ (𝑋 ∈ ({∅} × 𝐴) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)))
6		simprr 537	. . . . . 6 ⊢ ((𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {∅} ∧ (2^nd ‘𝑋) ∈ 𝐴)) → (2^nd ‘𝑋) ∈ 𝐴)
7	6	a1d 22	. . . . 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 121	. . . 4 ⊢ (𝑋 ∈ ({∅} × 𝐴) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
9		elxp6 6403	. . . . 5 ⊢ (𝑋 ∈ ({1_o} × 𝐵) ↔ (𝑋 = ⟨(1^st ‘𝑋), (2^nd ‘𝑋)⟩ ∧ ((1^st ‘𝑋) ∈ {1_o} ∧ (2^nd ‘𝑋) ∈ 𝐵)))
10		elsni 3727	. . . . . . 7 ⊢ ((1^st ‘𝑋) ∈ {1_o} → (1^st ‘𝑋) = 1_o)
11		1n0 6705	. . . . . . . 8 ⊢ 1_o ≠ ∅
12		neeq1 2433	. . . . . . . 8 ⊢ ((1^st ‘𝑋) = 1_o → ((1^st ‘𝑋) ≠ ∅ ↔ 1_o ≠ ∅))
13	11, 12	mpbiri 168	. . . . . . 7 ⊢ ((1^st ‘𝑋) = 1_o → (1^st ‘𝑋) ≠ ∅)
14		eqneqall 2430	. . . . . . . 8 ⊢ ((1^st ‘𝑋) = ∅ → ((1^st ‘𝑋) ≠ ∅ → (2^nd ‘𝑋) ∈ 𝐴))
15	14	com12 30	. . . . . . 7 ⊢ ((1^st ‘𝑋) ≠ ∅ → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
16	10, 13, 15	3syl 17	. . . . . 6 ⊢ ((1^st ‘𝑋) ∈ {1_o} → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
17	16	ad2antrl 494	. . . . 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 121	. . . 4 ⊢ (𝑋 ∈ ({1_o} × 𝐵) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
19	8, 18	jaoi 728	. . 3 ⊢ ((𝑋 ∈ ({∅} × 𝐴) ∨ 𝑋 ∈ ({1_o} × 𝐵)) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
20	4, 19	sylbi 121	. 2 ⊢ (𝑋 ∈ (𝐴 ⊔ 𝐵) → ((1^st ‘𝑋) = ∅ → (2^nd ‘𝑋) ∈ 𝐴))
21	20	imp 124	1 ⊢ ((𝑋 ∈ (𝐴 ⊔ 𝐵) ∧ (1^st ‘𝑋) = ∅) → (2^nd ‘𝑋) ∈ 𝐴)

Colors of variables: wff set class

This proof depends on syntax axioms: → wi 4 ∧ wa 104 ∨ wo 720 = wceq 1402 ∈ wcel 2209 ≠ wne 2420 ∪ cun 3218 ∅c0 3520 {csn 3709 ⟨cop 3712 × cxp 4772 ‘cfv 5377 1^st c1st 6372 2^nd c2nd 6373 1_oc1o 6680 ⊔ cdju 7377

This proof depends on axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-in1 623 ax-in2 624 ax-io 721 ax-5 1500 ax-7 1501 ax-gen 1502 ax-ie1 1546 ax-ie2 1547 ax-8 1557 ax-10 1558 ax-11 1559 ax-i12 1560 ax-bndl 1562 ax-4 1563 ax-17 1579 ax-i9 1583 ax-ial 1587 ax-i5r 1588 ax-14 2212 ax-ext 2220 ax-sep 4249 ax-nul 4259 ax-pow 4311 ax-pr 4346 ax-un 4578

This proof depends on definitions: df-bi 117 df-3an 1011 df-tru 1405 df-nf 1514 df-sb 1816 df-eu 2089 df-mo 2090 df-clab 2225 df-cleq 2231 df-clel 2234 df-nfc 2381 df-ne 2421 df-ral 2533 df-rex 2534 df-v 2823 df-sbc 3052 df-dif 3222 df-un 3224 df-in 3226 df-ss 3233 df-nul 3521 df-pw 3690 df-sn 3715 df-pr 3716 df-op 3718 df-uni 3936 df-br 4131 df-opab 4193 df-mpt 4194 df-id 4438 df-suc 4516 df-xp 4780 df-rel 4781 df-cnv 4782 df-co 4783 df-dm 4784 df-rn 4785 df-iota 5337 df-fun 5379 df-fv 5385 df-1st 6374 df-2nd 6375 df-1o 6687 df-dju 7378

This theorem is used by: updjudhf 7419 subctctexmid 17030

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