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| Mirrors > Home > ILE Home > Th. List > rextp | GIF version | ||
| Description: Convert a quantification over a triple to a disjunction. (Contributed by Mario Carneiro, 23-Apr-2015.) |
| Ref | Expression |
|---|---|
| raltp.1 | ⊢ 𝐴 ∈ V |
| raltp.2 | ⊢ 𝐵 ∈ V |
| raltp.3 | ⊢ 𝐶 ∈ V |
| raltp.4 | ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) |
| raltp.5 | ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒)) |
| raltp.6 | ⊢ (𝑥 = 𝐶 → (𝜑 ↔ 𝜃)) |
| Ref | Expression |
|---|---|
| rextp | ⊢ (∃𝑥 ∈ {𝐴, 𝐵, 𝐶}𝜑 ↔ (𝜓 ∨ 𝜒 ∨ 𝜃)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | raltp.1 | . 2 ⊢ 𝐴 ∈ V | |
| 2 | raltp.2 | . 2 ⊢ 𝐵 ∈ V | |
| 3 | raltp.3 | . 2 ⊢ 𝐶 ∈ V | |
| 4 | raltp.4 | . . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) | |
| 5 | raltp.5 | . . 3 ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒)) | |
| 6 | raltp.6 | . . 3 ⊢ (𝑥 = 𝐶 → (𝜑 ↔ 𝜃)) | |
| 7 | 4, 5, 6 | rextpg 3688 | . 2 ⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ V ∧ 𝐶 ∈ V) → (∃𝑥 ∈ {𝐴, 𝐵, 𝐶}𝜑 ↔ (𝜓 ∨ 𝜒 ∨ 𝜃))) |
| 8 | 1, 2, 3, 7 | mp3an 1350 | 1 ⊢ (∃𝑥 ∈ {𝐴, 𝐵, 𝐶}𝜑 ↔ (𝜓 ∨ 𝜒 ∨ 𝜃)) |
| Colors of variables: wff set class |
| Syntax hints: → wi 4 ↔ wb 105 ∨ w3o 980 = wceq 1373 ∈ wcel 2177 ∃wrex 2486 Vcvv 2773 {ctp 3636 |
| 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-io 711 ax-5 1471 ax-7 1472 ax-gen 1473 ax-ie1 1517 ax-ie2 1518 ax-8 1528 ax-10 1529 ax-11 1530 ax-i12 1531 ax-bndl 1533 ax-4 1534 ax-17 1550 ax-i9 1554 ax-ial 1558 ax-i5r 1559 ax-ext 2188 |
| This theorem depends on definitions: df-bi 117 df-3or 982 df-3an 983 df-tru 1376 df-nf 1485 df-sb 1787 df-clab 2193 df-cleq 2199 df-clel 2202 df-nfc 2338 df-rex 2491 df-v 2775 df-sbc 3000 df-un 3171 df-sn 3640 df-pr 3641 df-tp 3642 |
| This theorem is referenced by: (None) |
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