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Mirrors > Home > ILE Home > Th. List > elpr2 | GIF version |
Description: A member of an unordered pair of classes is one or the other of them. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 14-Oct-2005.) |
Ref | Expression |
---|---|
elpr2.1 | ⊢ 𝐵 ∈ V |
elpr2.2 | ⊢ 𝐶 ∈ V |
Ref | Expression |
---|---|
elpr2 | ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | elprg 3395 | . . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) | |
2 | 1 | ibi 165 | . 2 ⊢ (𝐴 ∈ {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)) |
3 | elpr2.1 | . . . . . 6 ⊢ 𝐵 ∈ V | |
4 | eleq1 2100 | . . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ∈ V ↔ 𝐵 ∈ V)) | |
5 | 3, 4 | mpbiri 157 | . . . . 5 ⊢ (𝐴 = 𝐵 → 𝐴 ∈ V) |
6 | elpr2.2 | . . . . . 6 ⊢ 𝐶 ∈ V | |
7 | eleq1 2100 | . . . . . 6 ⊢ (𝐴 = 𝐶 → (𝐴 ∈ V ↔ 𝐶 ∈ V)) | |
8 | 6, 7 | mpbiri 157 | . . . . 5 ⊢ (𝐴 = 𝐶 → 𝐴 ∈ V) |
9 | 5, 8 | jaoi 636 | . . . 4 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ V) |
10 | elprg 3395 | . . . 4 ⊢ (𝐴 ∈ V → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) | |
11 | 9, 10 | syl 14 | . . 3 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
12 | 11 | ibir 166 | . 2 ⊢ ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ {𝐵, 𝐶}) |
13 | 2, 12 | impbii 117 | 1 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 98 ∨ wo 629 = wceq 1243 ∈ wcel 1393 Vcvv 2557 {cpr 3376 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-io 630 ax-5 1336 ax-7 1337 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-10 1396 ax-11 1397 ax-i12 1398 ax-bndl 1399 ax-4 1400 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-i5r 1428 ax-ext 2022 |
This theorem depends on definitions: df-bi 110 df-tru 1246 df-nf 1350 df-sb 1646 df-clab 2027 df-cleq 2033 df-clel 2036 df-nfc 2167 df-v 2559 df-un 2922 df-sn 3381 df-pr 3382 |
This theorem is referenced by: elxr 8696 |
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