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Mirrors > Home > ILE Home > Th. List > injust | GIF version |
Description: Soundness justification theorem for df-in 2979. (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 2141 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴)) | |
2 | eleq1 2141 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵)) | |
3 | 1, 2 | anbi12d 456 | . . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵))) |
4 | 3 | cbvabv 2202 | . 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} |
5 | eleq1 2141 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴)) | |
6 | eleq1 2141 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)) | |
7 | 5, 6 | anbi12d 456 | . . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))) |
8 | 7 | cbvabv 2202 | . 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)} |
9 | 4, 8 | eqtri 2101 | 1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)} |
Colors of variables: wff set class |
Syntax hints: ∧ wa 102 = wceq 1284 ∈ wcel 1433 {cab 2067 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-io 662 ax-5 1376 ax-7 1377 ax-gen 1378 ax-ie1 1422 ax-ie2 1423 ax-8 1435 ax-10 1436 ax-11 1437 ax-i12 1438 ax-bndl 1439 ax-4 1440 ax-17 1459 ax-i9 1463 ax-ial 1467 ax-i5r 1468 ax-ext 2063 |
This theorem depends on definitions: df-bi 115 df-nf 1390 df-sb 1686 df-clab 2068 df-cleq 2074 df-clel 2077 |
This theorem is referenced by: (None) |
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