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Mirrors > Home > ILE Home > Th. List > nfnf | GIF version |
Description: If 𝑥 is not free in 𝜑, it is not free in Ⅎ𝑦𝜑. (Contributed by Mario Carneiro, 11-Aug-2016.) (Proof shortened by Wolf Lammen, 30-Dec-2017.) |
Ref | Expression |
---|---|
nfal.1 | ⊢ Ⅎ𝑥𝜑 |
Ref | Expression |
---|---|
nfnf | ⊢ Ⅎ𝑥Ⅎ𝑦𝜑 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-nf 1461 | . 2 ⊢ (Ⅎ𝑦𝜑 ↔ ∀𝑦(𝜑 → ∀𝑦𝜑)) | |
2 | nfal.1 | . . . 4 ⊢ Ⅎ𝑥𝜑 | |
3 | 2 | nfal 1576 | . . . 4 ⊢ Ⅎ𝑥∀𝑦𝜑 |
4 | 2, 3 | nfim 1572 | . . 3 ⊢ Ⅎ𝑥(𝜑 → ∀𝑦𝜑) |
5 | 4 | nfal 1576 | . 2 ⊢ Ⅎ𝑥∀𝑦(𝜑 → ∀𝑦𝜑) |
6 | 1, 5 | nfxfr 1474 | 1 ⊢ Ⅎ𝑥Ⅎ𝑦𝜑 |
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-7 1448 ax-gen 1449 ax-4 1510 ax-ial 1534 ax-i5r 1535 |
This theorem depends on definitions: df-bi 117 df-nf 1461 |
This theorem is referenced by: nfnfc 2326 |
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