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Mirrors > Home > MPE Home > Th. List > Mathboxes > disjif | Structured version Visualization version GIF version |
Description: Property of a disjoint collection: if 𝐵(𝑥) and 𝐵(𝑌) = 𝐷 have a common element 𝑍, then 𝑥 = 𝑌. (Contributed by Thierry Arnoux, 30-Dec-2016.) |
Ref | Expression |
---|---|
disjif.1 | ⊢ Ⅎ𝑥𝐶 |
disjif.2 | ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶) |
Ref | Expression |
---|---|
disjif | ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶)) → 𝑥 = 𝑌) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | inelcm 4414 | . 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅) | |
2 | disjif.1 | . . . . . 6 ⊢ Ⅎ𝑥𝐶 | |
3 | disjif.2 | . . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶) | |
4 | 2, 3 | disji2f 30321 | . . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅) |
5 | 4 | 3expia 1117 | . . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅)) |
6 | 5 | necon1d 3038 | . . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐵 ∩ 𝐶) ≠ ∅ → 𝑥 = 𝑌)) |
7 | 6 | 3impia 1113 | . 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐵 ∩ 𝐶) ≠ ∅) → 𝑥 = 𝑌) |
8 | 1, 7 | syl3an3 1161 | 1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶)) → 𝑥 = 𝑌) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ∧ wa 398 ∧ w3a 1083 = wceq 1533 ∈ wcel 2110 Ⅎwnfc 2961 ≠ wne 3016 ∩ cin 3935 ∅c0 4291 Disj wdisj 5024 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1907 ax-6 1966 ax-7 2011 ax-8 2112 ax-9 2120 ax-10 2141 ax-11 2156 ax-12 2172 ax-ext 2793 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3an 1085 df-tru 1536 df-ex 1777 df-nf 1781 df-sb 2066 df-mo 2618 df-eu 2650 df-clab 2800 df-cleq 2814 df-clel 2893 df-nfc 2963 df-ne 3017 df-ral 3143 df-rex 3144 df-reu 3145 df-rmo 3146 df-rab 3147 df-v 3497 df-sbc 3773 df-csb 3884 df-dif 3939 df-in 3943 df-nul 4292 df-disj 5025 |
This theorem is referenced by: disjabrex 30326 |
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