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Mirrors > Home > MPE Home > Th. List > unjust | Structured version Visualization version GIF version |
Description: Soundness justification theorem for df-un 3858. (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 2813 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴)) | |
2 | eleq1w 2813 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵)) | |
3 | 1, 2 | orbi12d 919 | . . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵))) |
4 | 3 | cbvabv 2804 | . 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} |
5 | eleq1w 2813 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴)) | |
6 | eleq1w 2813 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)) | |
7 | 5, 6 | orbi12d 919 | . . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵))) |
8 | 7 | cbvabv 2804 | . 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
9 | 4, 8 | eqtri 2759 | 1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
Colors of variables: wff setvar class |
Syntax hints: ∨ wo 847 = wceq 1543 ∈ wcel 2112 {cab 2714 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1803 ax-4 1817 ax-5 1918 ax-6 1976 ax-7 2018 ax-8 2114 ax-9 2122 ax-ext 2708 |
This theorem depends on definitions: df-bi 210 df-an 400 df-or 848 df-ex 1788 df-sb 2073 df-clab 2715 df-cleq 2728 df-clel 2809 |
This theorem is referenced by: (None) |
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