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Mirrors > Home > MPE Home > Th. List > unjust | Structured version Visualization version GIF version |
Description: Soundness justification theorem for df-un 3545. (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 2676 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐴 ↔ 𝑧 ∈ 𝐴)) | |
2 | eleq1 2676 | . . . 4 ⊢ (𝑥 = 𝑧 → (𝑥 ∈ 𝐵 ↔ 𝑧 ∈ 𝐵)) | |
3 | 1, 2 | orbi12d 742 | . . 3 ⊢ (𝑥 = 𝑧 → ((𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵) ↔ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵))) |
4 | 3 | cbvabv 2734 | . 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} |
5 | eleq1 2676 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴)) | |
6 | eleq1 2676 | . . . 4 ⊢ (𝑧 = 𝑦 → (𝑧 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)) | |
7 | 5, 6 | orbi12d 742 | . . 3 ⊢ (𝑧 = 𝑦 → ((𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵))) |
8 | 7 | cbvabv 2734 | . 2 ⊢ {𝑧 ∣ (𝑧 ∈ 𝐴 ∨ 𝑧 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
9 | 4, 8 | eqtri 2632 | 1 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)} |
Colors of variables: wff setvar class |
Syntax hints: ∨ wo 382 = wceq 1475 ∈ wcel 1977 {cab 2596 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 |
This theorem is referenced by: (None) |
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