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Mirrors > Home > MPE Home > Th. List > Mathboxes > equncomVD | Structured version Visualization version GIF version |
Description: If a class equals the union of two other classes, then it equals the union
of those two classes commuted. The following User's Proof is a Virtual
Deduction proof completed automatically by the tools program
completeusersproof.cmd, which invokes Mel L. O'Cat's mmj2 and Norm
Megill's Metamath Proof Assistant. equncom 4084 is equncomVD 42377 without
virtual deductions and was automatically derived from equncomVD 42377.
|
Ref | Expression |
---|---|
equncomVD | ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | idn1 42083 | . . . 4 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) | |
2 | uncom 4083 | . . . 4 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) | |
3 | eqeq1 2742 | . . . . 5 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = (𝐶 ∪ 𝐵) ↔ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵))) | |
4 | 3 | biimprd 247 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐶 ∪ 𝐵))) |
5 | 1, 2, 4 | e10 42203 | . . 3 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) |
6 | 5 | in1 42080 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → 𝐴 = (𝐶 ∪ 𝐵)) |
7 | idn1 42083 | . . . 4 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) | |
8 | eqeq2 2750 | . . . . 5 ⊢ ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵))) | |
9 | 8 | biimprcd 249 | . . . 4 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶))) |
10 | 7, 2, 9 | e10 42203 | . . 3 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) |
11 | 10 | in1 42080 | . 2 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶)) |
12 | 6, 11 | impbii 208 | 1 ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 205 = wceq 1539 ∪ cun 3881 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1799 ax-4 1813 ax-5 1914 ax-6 1972 ax-7 2012 ax-8 2110 ax-9 2118 ax-ext 2709 |
This theorem depends on definitions: df-bi 206 df-an 396 df-or 844 df-tru 1542 df-ex 1784 df-sb 2069 df-clab 2716 df-cleq 2730 df-clel 2817 df-v 3424 df-un 3888 df-vd1 42079 |
This theorem is referenced by: (None) |
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