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Mirrors > Home > ILE Home > Th. List > elnelall | GIF version |
Description: A contradiction concerning membership implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018.) |
Ref | Expression |
---|---|
elnelall | ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∉ 𝐵 → 𝜑)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-nel 2432 | . 2 ⊢ (𝐴 ∉ 𝐵 ↔ ¬ 𝐴 ∈ 𝐵) | |
2 | pm2.24 611 | . 2 ⊢ (𝐴 ∈ 𝐵 → (¬ 𝐴 ∈ 𝐵 → 𝜑)) | |
3 | 1, 2 | syl5bi 151 | 1 ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∉ 𝐵 → 𝜑)) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∈ wcel 2136 ∉ wnel 2431 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-in2 605 |
This theorem depends on definitions: df-bi 116 df-nel 2432 |
This theorem is referenced by: xnn0lenn0nn0 9801 |
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