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Theorem releldmdifi 8056

Description: One way of expressing membership in the difference of domains of two nested relations. (Contributed by AV, 26-Oct-2023.)

**Assertion**
Ref	Expression
releldmdifi	⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) → ∃𝑥 ∈ (𝐴 ∖ 𝐵)(1^st ‘𝑥) = 𝐶))

Distinct variable groups: 𝑥,𝐴 𝑥,𝐵 𝑥,𝐶

**Proof of Theorem releldmdifi**
Step	Hyp	Ref	Expression
1		eldif 3909	. . 3 ⊢ (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) ↔ (𝐶 ∈ dom 𝐴 ∧ ¬ 𝐶 ∈ dom 𝐵))
2		releldm2 8054	. . . . 5 ⊢ (Rel 𝐴 → (𝐶 ∈ dom 𝐴 ↔ ∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶))
3	2	adantr 486	. . . 4 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → (𝐶 ∈ dom 𝐴 ↔ ∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶))
4	3	anbi1d 643	. . 3 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → ((𝐶 ∈ dom 𝐴 ∧ ¬ 𝐶 ∈ dom 𝐵) ↔ (∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ¬ 𝐶 ∈ dom 𝐵)))
5	1, 4	bitrid 286	. 2 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) ↔ (∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ¬ 𝐶 ∈ dom 𝐵)))
6		simprl 783	. . . 4 ⊢ (((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) ∧ (∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ¬ 𝐶 ∈ dom 𝐵)) → ∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶)
7		relss 5758	. . . . . . . . . . . 12 ⊢ (𝐵 ⊆ 𝐴 → (Rel 𝐴 → Rel 𝐵))
8	7	impcom 413	. . . . . . . . . . 11 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → Rel 𝐵)
9		1stdm 8051	. . . . . . . . . . 11 ⊢ ((Rel 𝐵 ∧ 𝑥 ∈ 𝐵) → (1^st ‘𝑥) ∈ dom 𝐵)
10	8, 9	sylan 592	. . . . . . . . . 10 ⊢ (((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) ∧ 𝑥 ∈ 𝐵) → (1^st ‘𝑥) ∈ dom 𝐵)
11		eleq1 2849	. . . . . . . . . 10 ⊢ ((1^st ‘𝑥) = 𝐶 → ((1^st ‘𝑥) ∈ dom 𝐵 ↔ 𝐶 ∈ dom 𝐵))
12	10, 11	syl5ibcom 248	. . . . . . . . 9 ⊢ (((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) ∧ 𝑥 ∈ 𝐵) → ((1^st ‘𝑥) = 𝐶 → 𝐶 ∈ dom 𝐵))
13	12	rexlimdva 3164	. . . . . . . 8 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → (∃𝑥 ∈ 𝐵 (1^st ‘𝑥) = 𝐶 → 𝐶 ∈ dom 𝐵))
14	13	con3d 153	. . . . . . 7 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → (¬ 𝐶 ∈ dom 𝐵 → ¬ ∃𝑥 ∈ 𝐵 (1^st ‘𝑥) = 𝐶))
15		ralnex 3089	. . . . . . 7 ⊢ (∀𝑥 ∈ 𝐵 ¬ (1^st ‘𝑥) = 𝐶 ↔ ¬ ∃𝑥 ∈ 𝐵 (1^st ‘𝑥) = 𝐶)
16	14, 15	imbitrrdi 255	. . . . . 6 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → (¬ 𝐶 ∈ dom 𝐵 → ∀𝑥 ∈ 𝐵 ¬ (1^st ‘𝑥) = 𝐶))
17	16	adantld 496	. . . . 5 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → ((∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ¬ 𝐶 ∈ dom 𝐵) → ∀𝑥 ∈ 𝐵 ¬ (1^st ‘𝑥) = 𝐶))
18	17	imp 412	. . . 4 ⊢ (((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) ∧ (∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ¬ 𝐶 ∈ dom 𝐵)) → ∀𝑥 ∈ 𝐵 ¬ (1^st ‘𝑥) = 𝐶)
19		rexdifi 4097	. . . 4 ⊢ ((∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ∀𝑥 ∈ 𝐵 ¬ (1^st ‘𝑥) = 𝐶) → ∃𝑥 ∈ (𝐴 ∖ 𝐵)(1^st ‘𝑥) = 𝐶)
20	6, 18, 19	syl2anc 596	. . 3 ⊢ (((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) ∧ (∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ¬ 𝐶 ∈ dom 𝐵)) → ∃𝑥 ∈ (𝐴 ∖ 𝐵)(1^st ‘𝑥) = 𝐶)
21	20	ex 418	. 2 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → ((∃𝑥 ∈ 𝐴 (1^st ‘𝑥) = 𝐶 ∧ ¬ 𝐶 ∈ dom 𝐵) → ∃𝑥 ∈ (𝐴 ∖ 𝐵)(1^st ‘𝑥) = 𝐶))
22	5, 21	sylbid 243	1 ⊢ ((Rel 𝐴 ∧ 𝐵 ⊆ 𝐴) → (𝐶 ∈ (dom 𝐴 ∖ dom 𝐵) → ∃𝑥 ∈ (𝐴 ∖ 𝐵)(1^st ‘𝑥) = 𝐶))

Colors of variables: wff setvar class

This proof depends on syntax axioms: ¬ wn 3 → wi 4 ↔ wb 209 ∧ wa 401 = wceq 1570 ∈ wcel 2145 ∀wral 3077 ∃wrex 3087 ∖ cdif 3896 ⊆ wss 3899 dom cdm 5651 Rel wrel 5656 ‘cfv 6538 1^st c1st 7999

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 2733 ax-sep 5249 ax-nul 5260 ax-pr 5391 ax-un 7751

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 2565 df-eu 2595 df-clab 2740 df-cleq 2753 df-clel 2836 df-nfc 2910 df-ne 2957 df-ral 3078 df-rex 3088 df-rab 3414 df-v 3453 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-int 4908 df-br 5104 df-opab 5168 df-mpt 5187 df-id 5546 df-xp 5657 df-rel 5658 df-cnv 5659 df-co 5660 df-dm 5661 df-rn 5662 df-iota 6494 df-fun 6540 df-fv 6546 df-1st 8001 df-2nd 8002

This theorem is used by: funeldmdif 8059 satffunlem2lem2 36171

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