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Description: For any set x , there is a set not contained in x . The proof is based on Russell's paradox. (Contributed by NM, 23-Aug-1993) Remove use of ax-12 and ax-13 . (Revised by BJ, 31-May-2019) Extract from nalset . (Revised by Matthew House, 12-Apr-2026)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | exnelv | ⊢ ∃ 𝑦 ¬ 𝑦 ∈ 𝑥 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ax-sep | ⊢ ∃ 𝑦 ∀ 𝑧 ( 𝑧 ∈ 𝑦 ↔ ( 𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧 ) ) | |
| 2 | elequ1 | ⊢ ( 𝑧 = 𝑦 → ( 𝑧 ∈ 𝑥 ↔ 𝑦 ∈ 𝑥 ) ) | |
| 3 | elequ2 | ⊢ ( 𝑧 = 𝑦 → ( 𝑧 ∈ 𝑧 ↔ 𝑧 ∈ 𝑦 ) ) | |
| 4 | 3 | notbid | ⊢ ( 𝑧 = 𝑦 → ( ¬ 𝑧 ∈ 𝑧 ↔ ¬ 𝑧 ∈ 𝑦 ) ) |
| 5 | 2 4 | anbi12d | ⊢ ( 𝑧 = 𝑦 → ( ( 𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧 ) ↔ ( 𝑦 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑦 ) ) ) |
| 6 | 5 | bibi2d | ⊢ ( 𝑧 = 𝑦 → ( ( 𝑧 ∈ 𝑦 ↔ ( 𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧 ) ) ↔ ( 𝑧 ∈ 𝑦 ↔ ( 𝑦 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑦 ) ) ) ) |
| 7 | pclem6 | ⊢ ( ( 𝑧 ∈ 𝑦 ↔ ( 𝑦 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑦 ) ) → ¬ 𝑦 ∈ 𝑥 ) | |
| 8 | 6 7 | biimtrdi | ⊢ ( 𝑧 = 𝑦 → ( ( 𝑧 ∈ 𝑦 ↔ ( 𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧 ) ) → ¬ 𝑦 ∈ 𝑥 ) ) |
| 9 | 8 | spimvw | ⊢ ( ∀ 𝑧 ( 𝑧 ∈ 𝑦 ↔ ( 𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧 ) ) → ¬ 𝑦 ∈ 𝑥 ) |
| 10 | 1 9 | eximii | ⊢ ∃ 𝑦 ¬ 𝑦 ∈ 𝑥 |