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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 3984 | . 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅) | |
2 | disjif.1 | . . . . . 6 ⊢ Ⅎ𝑥𝐶 | |
3 | disjif.2 | . . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶) | |
4 | 2, 3 | disji2f 28772 | . . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅) |
5 | 4 | 3expia 1259 | . . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅)) |
6 | 5 | necon1d 2804 | . . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐵 ∩ 𝐶) ≠ ∅ → 𝑥 = 𝑌)) |
7 | 6 | 3impia 1253 | . 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐵 ∩ 𝐶) ≠ ∅) → 𝑥 = 𝑌) |
8 | 1, 7 | syl3an3 1353 | 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 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 Ⅎwnfc 2738 ≠ wne 2780 ∩ cin 3539 ∅c0 3874 Disj wdisj 4553 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-in 3547 df-nul 3875 df-disj 4554 |
This theorem is referenced by: disjabrex 28777 |
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