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Mirrors > Home > MPE Home > Th. List > nf5dh | Structured version Visualization version GIF version |
Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.) df-nf 1788 changed. (Revised by Wolf Lammen, 11-Oct-2021.) |
Ref | Expression |
---|---|
nf5dh.1 | ⊢ (𝜑 → ∀𝑥𝜑) |
nf5dh.2 | ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓)) |
Ref | Expression |
---|---|
nf5dh | ⊢ (𝜑 → Ⅎ𝑥𝜓) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nf5dh.1 | . . 3 ⊢ (𝜑 → ∀𝑥𝜑) | |
2 | nf5dh.2 | . . 3 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓)) | |
3 | 1, 2 | alrimih 1827 | . 2 ⊢ (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓)) |
4 | nf5-1 2143 | . 2 ⊢ (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓) | |
5 | 3, 4 | syl 17 | 1 ⊢ (𝜑 → Ⅎ𝑥𝜓) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ∀wal 1537 Ⅎwnf 1787 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1799 ax-4 1813 ax-10 2139 |
This theorem depends on definitions: df-bi 206 df-ex 1784 df-nf 1788 |
This theorem is referenced by: nf5dv 2146 hbimd 2298 ax12indalem 36886 ax12inda2ALT 36887 |
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