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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 4471 | . 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅) | |
2 | disjif.1 | . . . . . 6 ⊢ Ⅎ𝑥𝐶 | |
3 | disjif.2 | . . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶) | |
4 | 2, 3 | disji2f 32597 | . . . . 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 2960 | . . 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 395 ∧ w3a 1086 = wceq 1537 ∈ wcel 2106 Ⅎwnfc 2888 ≠ wne 2938 ∩ cin 3962 ∅c0 4339 Disj wdisj 5115 |
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 1908 ax-6 1965 ax-7 2005 ax-8 2108 ax-9 2116 ax-10 2139 ax-11 2155 ax-12 2175 ax-ext 2706 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3an 1088 df-tru 1540 df-fal 1550 df-ex 1777 df-nf 1781 df-sb 2063 df-mo 2538 df-clab 2713 df-cleq 2727 df-clel 2814 df-nfc 2890 df-ne 2939 df-ral 3060 df-rmo 3378 df-rab 3434 df-v 3480 df-sbc 3792 df-csb 3909 df-dif 3966 df-in 3970 df-nul 4340 df-disj 5116 |
This theorem is referenced by: disjabrex 32602 2ndresdju 32666 |
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