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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 3970 | . . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶)) | |
2 | 1 | com12 32 | . 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶)) |
3 | 2 | con3dimp 407 | 1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 394 ∈ wcel 2098 ⊆ wss 3944 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1905 ax-6 1963 ax-7 2003 ax-8 2100 |
This theorem depends on definitions: df-bi 206 df-an 395 df-ex 1774 df-clel 2802 df-ss 3961 |
This theorem is referenced by: nrelv 5802 ordtr3 6416 smndex2dnrinv 18875 frlmssuvc2 21746 1arithidom 33349 tfsconcatb0 42915 clsk1indlem1 43617 mapssbi 44725 fourierdlem10 45643 salgensscntex 45870 |
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