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| Mirrors > Home > NFE Home > Th. List > nfdh | GIF version | ||
| Description: Deduce that x is not free in ψ in a context. (Contributed by Mario Carneiro, 24-Sep-2016.) |
| Ref | Expression |
|---|---|
| nfdh.1 | ⊢ (φ → ∀xφ) |
| nfdh.2 | ⊢ (φ → (ψ → ∀xψ)) |
| Ref | Expression |
|---|---|
| nfdh | ⊢ (φ → Ⅎxψ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | nfdh.1 | . . 3 ⊢ (φ → ∀xφ) | |
| 2 | 1 | nfi 1551 | . 2 ⊢ Ⅎxφ |
| 3 | nfdh.2 | . 2 ⊢ (φ → (ψ → ∀xψ)) | |
| 4 | 2, 3 | nfd 1766 | 1 ⊢ (φ → Ⅎxψ) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ∀wal 1540 Ⅎwnf 1544 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-11 1746 |
| This theorem depends on definitions: df-bi 177 df-ex 1542 df-nf 1545 |
| This theorem is referenced by: hbimd 1815 ax11indalem 2197 ax11inda2ALT 2198 |
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