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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 4372 | . 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅) | |
2 | disjif.1 | . . . . . 6 ⊢ Ⅎ𝑥𝐶 | |
3 | disjif.2 | . . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶) | |
4 | 2, 3 | disji2f 30340 | . . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅) |
5 | 4 | 3expia 1118 | . . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅)) |
6 | 5 | necon1d 3009 | . . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐵 ∩ 𝐶) ≠ ∅ → 𝑥 = 𝑌)) |
7 | 6 | 3impia 1114 | . 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐵 ∩ 𝐶) ≠ ∅) → 𝑥 = 𝑌) |
8 | 1, 7 | syl3an3 1162 | 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 399 ∧ w3a 1084 = wceq 1538 ∈ wcel 2111 Ⅎwnfc 2936 ≠ wne 2987 ∩ cin 3880 ∅c0 4243 Disj wdisj 4995 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2770 |
This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2598 df-clab 2777 df-cleq 2791 df-clel 2870 df-nfc 2938 df-ne 2988 df-ral 3111 df-rmo 3114 df-rab 3115 df-v 3443 df-sbc 3721 df-csb 3829 df-dif 3884 df-in 3888 df-nul 4244 df-disj 4996 |
This theorem is referenced by: disjabrex 30345 2ndresdju 30411 |
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