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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 4400 | . 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅) | |
2 | disjif.1 | . . . . . 6 ⊢ Ⅎ𝑥𝐶 | |
3 | disjif.2 | . . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶) | |
4 | 2, 3 | disji2f 30913 | . . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅) |
5 | 4 | 3expia 1120 | . . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅)) |
6 | 5 | necon1d 2965 | . . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐵 ∩ 𝐶) ≠ ∅ → 𝑥 = 𝑌)) |
7 | 6 | 3impia 1116 | . 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐵 ∩ 𝐶) ≠ ∅) → 𝑥 = 𝑌) |
8 | 1, 7 | syl3an3 1164 | 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 396 ∧ w3a 1086 = wceq 1539 ∈ wcel 2106 Ⅎwnfc 2887 ≠ wne 2943 ∩ cin 3887 ∅c0 4258 Disj wdisj 5041 |
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 1913 ax-6 1971 ax-7 2011 ax-8 2108 ax-9 2116 ax-10 2137 ax-11 2154 ax-12 2171 ax-ext 2709 |
This theorem depends on definitions: df-bi 206 df-an 397 df-or 845 df-3an 1088 df-tru 1542 df-fal 1552 df-ex 1783 df-nf 1787 df-sb 2068 df-mo 2540 df-clab 2716 df-cleq 2730 df-clel 2816 df-nfc 2889 df-ne 2944 df-ral 3069 df-rmo 3071 df-rab 3073 df-v 3433 df-sbc 3718 df-csb 3834 df-dif 3891 df-in 3895 df-nul 4259 df-disj 5042 |
This theorem is referenced by: disjabrex 30918 2ndresdju 30983 |
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