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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 3096 | . . 3 ⊢ (𝐵 ⊆ 𝐶 → (𝐴 ∈ 𝐵 → 𝐴 ∈ 𝐶)) | |
2 | 1 | com12 30 | . 2 ⊢ (𝐴 ∈ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ∈ 𝐶)) |
3 | 2 | con3dimp 625 | 1 ⊢ ((𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶) → ¬ 𝐵 ⊆ 𝐶) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 103 ∈ wcel 1481 ⊆ wss 3076 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 604 ax-in2 605 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1483 ax-11 1485 ax-4 1488 ax-17 1507 ax-i9 1511 ax-ial 1515 ax-i5r 1516 ax-ext 2122 |
This theorem depends on definitions: df-bi 116 df-nf 1438 df-sb 1737 df-clab 2127 df-cleq 2133 df-clel 2136 df-in 3082 df-ss 3089 |
This theorem is referenced by: (None) |
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