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Mirrors > Home > MPE Home > Th. List > nfnan | Structured version Visualization version GIF version |
Description: If 𝑥 is not free in 𝜑 and 𝜓, then it is not free in (𝜑 ⊼ 𝜓). (Contributed by Scott Fenton, 2-Jan-2018.) |
Ref | Expression |
---|---|
nfan.1 | ⊢ Ⅎ𝑥𝜑 |
nfan.2 | ⊢ Ⅎ𝑥𝜓 |
Ref | Expression |
---|---|
nfnan | ⊢ Ⅎ𝑥(𝜑 ⊼ 𝜓) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-nan 1487 | . 2 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓)) | |
2 | nfan.1 | . . . 4 ⊢ Ⅎ𝑥𝜑 | |
3 | nfan.2 | . . . 4 ⊢ Ⅎ𝑥𝜓 | |
4 | 2, 3 | nfan 1902 | . . 3 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜓) |
5 | 4 | nfn 1860 | . 2 ⊢ Ⅎ𝑥 ¬ (𝜑 ∧ 𝜓) |
6 | 1, 5 | nfxfr 1855 | 1 ⊢ Ⅎ𝑥(𝜑 ⊼ 𝜓) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 ∧ wa 396 ⊼ wnan 1486 Ⅎwnf 1786 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 |
This theorem depends on definitions: df-bi 206 df-an 397 df-or 845 df-nan 1487 df-tru 1542 df-ex 1783 df-nf 1787 |
This theorem is referenced by: (None) |
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