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| Mirrors > Home > ILE Home > Th. List > nelss | 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 3186 | . . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶)) | |
| 2 | 1 | com12 30 | . 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶)) |
| 3 | 2 | con3dimp 636 | 1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶) |
| Colors of variables: wff set class |
| Syntax hints: ¬ wn 3 → wi 4 ∧ wa 104 ∈ wcel 2175 ⊆ wss 3165 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-in1 615 ax-in2 616 ax-5 1469 ax-7 1470 ax-gen 1471 ax-ie1 1515 ax-ie2 1516 ax-8 1526 ax-11 1528 ax-4 1532 ax-17 1548 ax-i9 1552 ax-ial 1556 ax-i5r 1557 ax-ext 2186 |
| This theorem depends on definitions: df-bi 117 df-nf 1483 df-sb 1785 df-clab 2191 df-cleq 2197 df-clel 2200 df-in 3171 df-ss 3178 |
| This theorem is referenced by: (None) |
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