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Mirrors > Home > MPE Home > Th. List > nf3or | Structured version Visualization version GIF version |
Description: If 𝑥 is not free in 𝜑, 𝜓, and 𝜒, it is not free in (𝜑 ∨ 𝜓 ∨ 𝜒). (Contributed by Mario Carneiro, 11-Aug-2016.) |
Ref | Expression |
---|---|
nf.1 | ⊢ Ⅎ𝑥𝜑 |
nf.2 | ⊢ Ⅎ𝑥𝜓 |
nf.3 | ⊢ Ⅎ𝑥𝜒 |
Ref | Expression |
---|---|
nf3or | ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓 ∨ 𝜒) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-3or 1032 | . 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒)) | |
2 | nf.1 | . . . 4 ⊢ Ⅎ𝑥𝜑 | |
3 | nf.2 | . . . 4 ⊢ Ⅎ𝑥𝜓 | |
4 | 2, 3 | nfor 1822 | . . 3 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓) |
5 | nf.3 | . . 3 ⊢ Ⅎ𝑥𝜒 | |
6 | 4, 5 | nfor 1822 | . 2 ⊢ Ⅎ𝑥((𝜑 ∨ 𝜓) ∨ 𝜒) |
7 | 1, 6 | nfxfr 1771 | 1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓 ∨ 𝜒) |
Colors of variables: wff setvar class |
Syntax hints: ∨ wo 382 ∨ w3o 1030 Ⅎwnf 1699 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 |
This theorem depends on definitions: df-bi 196 df-or 384 df-3or 1032 df-tru 1478 df-ex 1696 df-nf 1701 |
This theorem is referenced by: nfso 4965 |
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