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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 3954 is equncomVD 39852 without
virtual deductions and was automatically derived from equncomVD 39852.
|
Ref | Expression |
---|---|
equncomVD | ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | idn1 39548 | . . . 4 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) | |
2 | uncom 3953 | . . . 4 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) | |
3 | eqeq1 2801 | . . . . 5 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = (𝐶 ∪ 𝐵) ↔ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵))) | |
4 | 3 | biimprd 240 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐶 ∪ 𝐵))) |
5 | 1, 2, 4 | e10 39677 | . . 3 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) |
6 | 5 | in1 39545 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → 𝐴 = (𝐶 ∪ 𝐵)) |
7 | idn1 39548 | . . . 4 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) | |
8 | eqeq2 2808 | . . . . 5 ⊢ ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵))) | |
9 | 8 | biimprcd 242 | . . . 4 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶))) |
10 | 7, 2, 9 | e10 39677 | . . 3 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) |
11 | 10 | in1 39545 | . 2 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶)) |
12 | 6, 11 | impbii 201 | 1 ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 198 = wceq 1653 ∪ cun 3765 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1891 ax-4 1905 ax-5 2006 ax-6 2072 ax-7 2107 ax-9 2166 ax-10 2185 ax-11 2200 ax-12 2213 ax-ext 2775 |
This theorem depends on definitions: df-bi 199 df-an 386 df-or 875 df-tru 1657 df-ex 1876 df-nf 1880 df-sb 2065 df-clab 2784 df-cleq 2790 df-clel 2793 df-nfc 2928 df-v 3385 df-un 3772 df-vd1 39544 |
This theorem is referenced by: (None) |
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