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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 4130 is equncomVD 41195 without
virtual deductions and was automatically derived from equncomVD 41195.
|
Ref | Expression |
---|---|
equncomVD | ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | idn1 40901 | . . . 4 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) | |
2 | uncom 4129 | . . . 4 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) | |
3 | eqeq1 2825 | . . . . 5 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = (𝐶 ∪ 𝐵) ↔ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵))) | |
4 | 3 | biimprd 250 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐶 ∪ 𝐵))) |
5 | 1, 2, 4 | e10 41021 | . . 3 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) |
6 | 5 | in1 40898 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → 𝐴 = (𝐶 ∪ 𝐵)) |
7 | idn1 40901 | . . . 4 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) | |
8 | eqeq2 2833 | . . . . 5 ⊢ ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵))) | |
9 | 8 | biimprcd 252 | . . . 4 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → ((𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶))) |
10 | 7, 2, 9 | e10 41021 | . . 3 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) |
11 | 10 | in1 40898 | . 2 ⊢ (𝐴 = (𝐶 ∪ 𝐵) → 𝐴 = (𝐵 ∪ 𝐶)) |
12 | 6, 11 | impbii 211 | 1 ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 208 = wceq 1533 ∪ cun 3934 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1907 ax-6 1966 ax-7 2011 ax-8 2112 ax-9 2120 ax-10 2141 ax-11 2156 ax-12 2172 ax-ext 2793 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-tru 1536 df-ex 1777 df-nf 1781 df-sb 2066 df-clab 2800 df-cleq 2814 df-clel 2893 df-nfc 2963 df-v 3497 df-un 3941 df-vd1 40897 |
This theorem is referenced by: (None) |
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