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| Mirrors > Home > ILE Home > Th. List > nelrdva | GIF version | ||
| Description: Deduce negative membership from an implication. (Contributed by Thierry Arnoux, 27-Nov-2017.) |
| Ref | Expression |
|---|---|
| nelrdva.1 | ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵) |
| Ref | Expression |
|---|---|
| nelrdva | ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | eqidd 2206 | . 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 = 𝐵) | |
| 2 | eleq1 2268 | . . . . . . 7 ⊢ (𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴)) | |
| 3 | 2 | anbi2d 464 | . . . . . 6 ⊢ (𝑥 = 𝐵 → ((𝜑 ∧ 𝑥 ∈ 𝐴) ↔ (𝜑 ∧ 𝐵 ∈ 𝐴))) |
| 4 | neeq1 2389 | . . . . . 6 ⊢ (𝑥 = 𝐵 → (𝑥 ≠ 𝐵 ↔ 𝐵 ≠ 𝐵)) | |
| 5 | 3, 4 | imbi12d 234 | . . . . 5 ⊢ (𝑥 = 𝐵 → (((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵) ↔ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵))) |
| 6 | nelrdva.1 | . . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 ≠ 𝐵) | |
| 7 | 5, 6 | vtoclg 2833 | . . . 4 ⊢ (𝐵 ∈ 𝐴 → ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵)) |
| 8 | 7 | anabsi7 581 | . . 3 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → 𝐵 ≠ 𝐵) |
| 9 | 8 | neneqd 2397 | . 2 ⊢ ((𝜑 ∧ 𝐵 ∈ 𝐴) → ¬ 𝐵 = 𝐵) |
| 10 | 1, 9 | pm2.65da 663 | 1 ⊢ (𝜑 → ¬ 𝐵 ∈ 𝐴) |
| Colors of variables: wff set class |
| Syntax hints: ¬ wn 3 → wi 4 ∧ wa 104 = wceq 1373 ∈ wcel 2176 ≠ wne 2376 |
| 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-in1 615 ax-in2 616 ax-io 711 ax-5 1470 ax-7 1471 ax-gen 1472 ax-ie1 1516 ax-ie2 1517 ax-8 1527 ax-10 1528 ax-11 1529 ax-i12 1530 ax-bndl 1532 ax-4 1533 ax-17 1549 ax-i9 1553 ax-ial 1557 ax-i5r 1558 ax-ext 2187 |
| This theorem depends on definitions: df-bi 117 df-tru 1376 df-nf 1484 df-sb 1786 df-clab 2192 df-cleq 2198 df-clel 2201 df-nfc 2337 df-ne 2377 df-v 2774 |
| This theorem is referenced by: (None) |
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