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Mirrors > Home > ILE Home > Th. List > injust | GIF version |
Description: Soundness justification theorem for df-in 3082. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.) |
Ref | Expression |
---|---|
injust | ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)} |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eleq1 2203 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴)) | |
2 | eleq1 2203 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵)) | |
3 | 1, 2 | anbi12d 465 | . . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))) |
4 | 3 | cbvabv 2265 | . 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} |
5 | eleq1 2203 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴)) | |
6 | eleq1 2203 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)) | |
7 | 5, 6 | anbi12d 465 | . . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))) |
8 | 7 | cbvabv 2265 | . 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)} |
9 | 4, 8 | eqtri 2161 | 1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)} |
Colors of variables: wff set class |
Syntax hints: ∧ wa 103 = wceq 1332 ∈ wcel 1481 {cab 2126 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1483 ax-10 1484 ax-11 1485 ax-i12 1486 ax-bndl 1487 ax-4 1488 ax-17 1507 ax-i9 1511 ax-ial 1515 ax-i5r 1516 ax-ext 2122 |
This theorem depends on definitions: df-bi 116 df-nf 1438 df-sb 1737 df-clab 2127 df-cleq 2133 df-clel 2136 |
This theorem is referenced by: (None) |
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