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Mirrors > Home > MPE Home > Th. List > nelss | Structured version Visualization version GIF version |
Description: Demonstrate by witnesses that two classes lack a subclass relation. (Contributed by Stefan O'Rear, 5-Feb-2015.) |
Ref | Expression |
---|---|
nelss | ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | ssel 3960 | . . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶)) | |
2 | 1 | com12 32 | . 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶)) |
3 | 2 | con3dimp 409 | 1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 396 ∈ wcel 2105 ⊆ wss 3935 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1787 ax-4 1801 ax-5 1902 ax-6 1961 ax-7 2006 ax-8 2107 ax-9 2115 ax-10 2136 ax-11 2151 ax-12 2167 ax-ext 2793 |
This theorem depends on definitions: df-bi 208 df-an 397 df-or 842 df-tru 1531 df-ex 1772 df-nf 1776 df-sb 2061 df-clab 2800 df-cleq 2814 df-clel 2893 df-in 3942 df-ss 3951 |
This theorem is referenced by: nrelv 5667 ordtr3 6230 frlmssuvc2 20869 clsk1indlem1 40275 mapssbi 41356 fourierdlem10 42283 salgensscntex 42508 smndex2dnrinv 43985 |
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