| Metamath Proof Explorer |
< Previous
Next >
Related theorems GIF version |
| Description: A rearrangement of intersection. |
| Ref | Expression |
|---|---|
| in12 | ⊢ (A ∩ (B ∩ C)) = (B ∩ (A ∩ C)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | incom 1636 | . . 3 ⊢ (A ∩ B) = (B ∩ A) | |
| 2 | 1 | ineq1i 1641 | . 2 ⊢ ((A ∩ B) ∩ C) = ((B ∩ A) ∩ C) |
| 3 | inass 1650 | . 2 ⊢ ((A ∩ B) ∩ C) = (A ∩ (B ∩ C)) | |
| 4 | inass 1650 | . 2 ⊢ ((B ∩ A) ∩ C) = (B ∩ (A ∩ C)) | |
| 5 | 2, 3, 4 | 3eqtr3 1124 | 1 ⊢ (A ∩ (B ∩ C)) = (B ∩ (A ∩ C)) |
| Colors of variables: wff set class |
| Syntax hints: = wceq 1091 ∩ cin 1486 |
| This theorem is referenced by: in4 1653 kmlem11 3590 fh1 5518 fh2 5519 |
| This theorem was proved from axioms: ax-1 3 ax-2 4 ax-3 5 ax-mp 6 ax-4 673 ax-5 674 ax-6 675 ax-7 676 ax-gen 677 ax-8 798 ax-9 799 ax-10 800 ax-11 801 ax-12 802 ax-16 922 ax-17 925 ax-ext 1074 |
| This theorem depends on definitions: df-bi 128 df-an 198 df-ex 679 df-sb 853 df-clab 1093 df-cleq 1097 df-clel 1099 df-v 1349 df-in 1491 |