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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 4182 is equncomVD 44839 without
virtual deductions and was automatically derived from equncomVD 44839.
|
Ref | Expression |
---|---|
equncomVD | ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | idn1 44545 | . . . 4 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) | |
2 | uncom 4181 | . . . 4 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵) | |
3 | eqeq1 2744 | . . . . 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 44665 | . . 3 ⊢ ( 𝐴 = (𝐵 ∪ 𝐶) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) |
6 | 5 | in1 44542 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → 𝐴 = (𝐶 ∪ 𝐵)) |
7 | idn1 44545 | . . . 4 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐶 ∪ 𝐵) ) | |
8 | eqeq2 2752 | . . . . 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 44665 | . . 3 ⊢ ( 𝐴 = (𝐶 ∪ 𝐵) ▶ 𝐴 = (𝐵 ∪ 𝐶) ) |
11 | 10 | in1 44542 | . 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 1537 ∪ cun 3974 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1793 ax-4 1807 ax-5 1909 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-ext 2711 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 847 df-tru 1540 df-ex 1778 df-sb 2065 df-clab 2718 df-cleq 2732 df-clel 2819 df-v 3490 df-un 3981 df-vd1 44541 |
This theorem is referenced by: (None) |
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