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Mirrors > Home > ILE Home > Th. List > nfbid | GIF version |
Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, it is not free in (𝜓 ↔ 𝜒). (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 29-Dec-2017.) |
Ref | Expression |
---|---|
nfbid.1 | ⊢ (𝜑 → Ⅎ𝑥𝜓) |
nfbid.2 | ⊢ (𝜑 → Ⅎ𝑥𝜒) |
Ref | Expression |
---|---|
nfbid | ⊢ (𝜑 → Ⅎ𝑥(𝜓 ↔ 𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | dfbi2 380 | . 2 ⊢ ((𝜓 ↔ 𝜒) ↔ ((𝜓 → 𝜒) ∧ (𝜒 → 𝜓))) | |
2 | nfbid.1 | . . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓) | |
3 | nfbid.2 | . . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜒) | |
4 | 2, 3 | nfimd 1517 | . . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → 𝜒)) |
5 | 3, 2 | nfimd 1517 | . . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜒 → 𝜓)) |
6 | 4, 5 | nfand 1500 | . 2 ⊢ (𝜑 → Ⅎ𝑥((𝜓 → 𝜒) ∧ (𝜒 → 𝜓))) |
7 | 1, 6 | nfxfrd 1404 | 1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ↔ 𝜒)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 102 ↔ wb 103 Ⅎwnf 1389 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-5 1376 ax-gen 1378 ax-4 1440 ax-ial 1467 ax-i5r 1468 |
This theorem depends on definitions: df-bi 115 df-nf 1390 |
This theorem is referenced by: nfbi 1521 nfeudv 1956 nfeqd 2233 nfiotadxy 4890 iota2df 4911 bdsepnft 10678 strcollnft 10779 |
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