![]() |
Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
|
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 3814 | . . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶)) | |
2 | 1 | com12 32 | . 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶)) |
3 | 2 | con3dimp 399 | 1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 386 ∈ wcel 2106 ⊆ wss 3791 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1839 ax-4 1853 ax-5 1953 ax-6 2021 ax-7 2054 ax-9 2115 ax-10 2134 ax-11 2149 ax-12 2162 ax-ext 2753 |
This theorem depends on definitions: df-bi 199 df-an 387 df-or 837 df-tru 1605 df-ex 1824 df-nf 1828 df-sb 2012 df-clab 2763 df-cleq 2769 df-clel 2773 df-in 3798 df-ss 3805 |
This theorem is referenced by: nrelv 5471 ordtr3 6021 frlmssuvc2 20538 clsk1indlem1 39292 mapssbi 40319 fourierdlem10 41254 salgensscntex 41479 |
Copyright terms: Public domain | W3C validator |