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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 3909 | . . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶)) | |
2 | 1 | com12 32 | . 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶)) |
3 | 2 | con3dimp 412 | 1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 399 ∈ wcel 2112 ⊆ wss 3882 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1803 ax-4 1817 ax-5 1918 ax-6 1976 ax-7 2016 ax-8 2114 ax-9 2122 ax-ext 2710 |
This theorem depends on definitions: df-bi 210 df-an 400 df-tru 1546 df-ex 1788 df-sb 2073 df-clab 2717 df-cleq 2731 df-clel 2818 df-v 3424 df-in 3889 df-ss 3899 |
This theorem is referenced by: nrelv 5687 ordtr3 6278 smndex2dnrinv 18372 frlmssuvc2 20787 clsk1indlem1 41367 mapssbi 42461 fourierdlem10 43366 salgensscntex 43591 |
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