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Mirrors > Home > ILE 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 1462 | . 2 ⊢ Ⅎ𝑥𝜑 |
3 | nfdh.2 | . 2 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓)) | |
4 | 2, 3 | nfd 1523 | 1 ⊢ (𝜑 → Ⅎ𝑥𝜓) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∀wal 1351 Ⅎwnf 1460 |
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-5 1447 ax-gen 1449 ax-4 1510 |
This theorem depends on definitions: df-bi 117 df-nf 1461 |
This theorem is referenced by: hbsbd 1982 |
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