| Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
| Mirrors > Home > MPE Home > Th. List > unjust | Structured version Visualization version GIF version | ||
| Description: Soundness justification theorem for df-un 3927. (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 | eleq1w 2812 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴)) | |
| 2 | eleq1w 2812 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵)) | |
| 3 | 1, 2 | orbi12d 918 | . . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵))) |
| 4 | 3 | cbvabv 2800 | . 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} |
| 5 | eleq1w 2812 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴)) | |
| 6 | eleq1w 2812 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)) | |
| 7 | 5, 6 | orbi12d 918 | . . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵))) |
| 8 | 7 | cbvabv 2800 | . 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
| 9 | 4, 8 | eqtri 2753 | 1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
| Colors of variables: wff setvar class |
| Syntax hints: ∨ wo 847 = wceq 1540 ∈ wcel 2109 {cab 2708 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2008 ax-8 2111 ax-9 2119 ax-ext 2702 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-ex 1780 df-sb 2066 df-clab 2709 df-cleq 2722 df-clel 2804 |
| This theorem is referenced by: (None) |
| Copyright terms: Public domain | W3C validator |