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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 4168 is equncomVD 44865 without
virtual deductions and was automatically derived from equncomVD 44865.
|
Ref | Expression |
---|---|
equncomVD | ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | idn1 44571 | . . . 4 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) | |
2 | uncom 4167 | . . . 4 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) | |
3 | eqeq1 2738 | . . . . 5 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = (𝐶 ∪ 𝐵) ↔ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵))) | |
4 | 3 | biimprd 248 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐶 ∪ 𝐵))) |
5 | 1, 2, 4 | e10 44691 | . . 3 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) |
6 | 5 | in1 44568 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → 𝐴 = (𝐶 ∪ 𝐵)) |
7 | idn1 44571 | . . . 4 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) | |
8 | eqeq2 2746 | . . . . 5 ⊢ ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵))) | |
9 | 8 | biimprcd 250 | . . . 4 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶))) |
10 | 7, 2, 9 | e10 44691 | . . 3 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) |
11 | 10 | in1 44568 | . 2 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶)) |
12 | 6, 11 | impbii 209 | 1 ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 206 = wceq 1536 ∪ cun 3960 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1791 ax-4 1805 ax-5 1907 ax-6 1964 ax-7 2004 ax-8 2107 ax-9 2115 ax-ext 2705 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-tru 1539 df-ex 1776 df-sb 2062 df-clab 2712 df-cleq 2726 df-clel 2813 df-v 3479 df-un 3967 df-vd1 44567 |
This theorem is referenced by: (None) |
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