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| Mirrors > Home > MPE Home > Th. List > nfd | Structured version Visualization version GIF version | ||
| Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Wolf Lammen, 16-Sep-2021.) |
| Ref | Expression |
|---|---|
| nfd.1 | ⊢ (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓)) |
| Ref | Expression |
|---|---|
| nfd | ⊢ (𝜑 → Ⅎ𝑥𝜓) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | nfd.1 | . 2 ⊢ (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓)) | |
| 2 | df-nf 1791 | . 2 ⊢ (Ⅎ𝑥𝜓 ↔ (∃𝑥𝜓 → ∀𝑥𝜓)) | |
| 3 | 1, 2 | sylibr 235 | 1 ⊢ (𝜑 → Ⅎ𝑥𝜓) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ∀wal 1545 ∃wex 1786 Ⅎwnf 1790 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 |
| This theorem depends on definitions: df-bi 208 df-nf 1791 |
| This theorem is referenced by: nftht 1799 nfntht 1800 nfimd 1901 nf5-1 2156 axc16nf 2275 nfald 2337 nfeqf2 2385 bj-nfald 37495 |
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