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Mirrors > Home > MPE Home > Th. List > elpr2g | Structured version Visualization version GIF version |
Description: A member of a pair of sets is one or the other of them, and conversely. Exercise 1 of [TakeutiZaring] p. 15. (Contributed by NM, 14-Oct-2005.) Generalize from sethood hypothesis to sethood antecedent. (Revised by BJ, 25-May-2024.) |
Ref | Expression |
---|---|
elpr2g | ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | elex 3466 | . . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} → 𝐴 ∈ V) | |
2 | 1 | a1i 11 | . 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} → 𝐴 ∈ V)) |
3 | elex 3466 | . . . 4 ⊢ (𝐵 ∈ 𝑉 → 𝐵 ∈ V) | |
4 | eleq1a 2833 | . . . 4 ⊢ (𝐵 ∈ V → (𝐴 = 𝐵 → 𝐴 ∈ V)) | |
5 | 3, 4 | syl 17 | . . 3 ⊢ (𝐵 ∈ 𝑉 → (𝐴 = 𝐵 → 𝐴 ∈ V)) |
6 | elex 3466 | . . . 4 ⊢ (𝐶 ∈ 𝑊 → 𝐶 ∈ V) | |
7 | eleq1a 2833 | . . . 4 ⊢ (𝐶 ∈ V → (𝐴 = 𝐶 → 𝐴 ∈ V)) | |
8 | 6, 7 | syl 17 | . . 3 ⊢ (𝐶 ∈ 𝑊 → (𝐴 = 𝐶 → 𝐴 ∈ V)) |
9 | 5, 8 | jaao 954 | . 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → ((𝐴 = 𝐵 ∨ 𝐴 = 𝐶) → 𝐴 ∈ V)) |
10 | elprg 4612 | . . 3 ⊢ (𝐴 ∈ V → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) | |
11 | 10 | a1i 11 | . 2 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ V → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)))) |
12 | 2, 9, 11 | pm5.21ndd 381 | 1 ⊢ ((𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑊) → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 205 ∧ wa 397 ∨ wo 846 = wceq 1542 ∈ wcel 2107 Vcvv 3448 {cpr 4593 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1914 ax-6 1972 ax-7 2012 ax-8 2109 ax-9 2117 ax-ext 2708 |
This theorem depends on definitions: df-bi 206 df-an 398 df-or 847 df-tru 1545 df-ex 1783 df-sb 2069 df-clab 2715 df-cleq 2729 df-clel 2815 df-v 3450 df-un 3920 df-sn 4592 df-pr 4594 |
This theorem is referenced by: elpr2 4616 |
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