| Intuitionistic Logic Explorer |
< Previous
Next >
Nearby theorems |
||
| Mirrors > Home > ILE Home > Th. List > unjust | GIF version | ||
| Description: Soundness justification theorem for df-un 3174. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.) |
| Ref | Expression |
|---|---|
| unjust | ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | eleq1 2269 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴)) | |
| 2 | eleq1 2269 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵)) | |
| 3 | 1, 2 | orbi12d 795 | . . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵))) |
| 4 | 3 | cbvabv 2331 | . 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} |
| 5 | eleq1 2269 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴)) | |
| 6 | eleq1 2269 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)) | |
| 7 | 5, 6 | orbi12d 795 | . . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵))) |
| 8 | 7 | cbvabv 2331 | . 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
| 9 | 4, 8 | eqtri 2227 | 1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
| Colors of variables: wff set class |
| Syntax hints: ∨ wo 710 = wceq 1373 ∈ wcel 2177 {cab 2192 |
| 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-nf 1485 df-sb 1787 df-clab 2193 df-cleq 2199 df-clel 2202 |
| This theorem is referenced by: (None) |
| Copyright terms: Public domain | W3C validator |