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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 3908 | . . 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 2111 ⊆ wss 3881 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-ext 2770 |
This theorem depends on definitions: df-bi 210 df-an 400 df-ex 1782 df-sb 2070 df-clab 2777 df-cleq 2791 df-clel 2870 df-v 3443 df-in 3888 df-ss 3898 |
This theorem is referenced by: nrelv 5637 ordtr3 6204 smndex2dnrinv 18072 frlmssuvc2 20484 clsk1indlem1 40748 mapssbi 41842 fourierdlem10 42759 salgensscntex 42984 |
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