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| Mirrors > Home > ILE Home > Th. List > nfd2 | GIF version | ||
| Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Wolf Lammen, 16-Sep-2021.) |
| Ref | Expression |
|---|---|
| nfd2.1 | ⊢ (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓)) |
| Ref | Expression |
|---|---|
| nfd2 | ⊢ (𝜑 → Ⅎ𝑥𝜓) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | nfd2.1 | . 2 ⊢ (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓)) | |
| 2 | nf2 1690 | . 2 ⊢ (Ⅎ𝑥𝜓 ↔ (∃𝑥𝜓 → ∀𝑥𝜓)) | |
| 3 | 1, 2 | sylibr 134 | 1 ⊢ (𝜑 → Ⅎ𝑥𝜓) |
| Colors of variables: wff set class |
| Syntax hints: → wi 4 ∀wal 1370 Ⅎwnf 1482 ∃wex 1514 |
| 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-gen 1471 ax-ie2 1516 ax-4 1532 ax-ial 1556 |
| This theorem depends on definitions: df-bi 117 df-nf 1483 |
| This theorem is referenced by: nf5-1 2051 |
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