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| Mirrors > Home > QLE Home > Th. List > vneulem9 | GIF version | ||
| Description: Part of von Neumann's lemma. Lemma 9, Kalmbach p. 96. (Contributed by NM, 31-Mar-2011.) |
| Ref | Expression |
|---|---|
| vneulem6.1 | ((a ∪ b) ∩ (c ∪ d)) = 0 |
| Ref | Expression |
|---|---|
| vneulem9 | (((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = ((c ∩ d) ∪ (a ∪ b)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ancom 74 | . . 3 (((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = (((a ∪ b) ∪ c) ∩ ((a ∪ b) ∪ d)) | |
| 2 | vneulem5 1135 | . . 3 (((a ∪ b) ∪ c) ∩ ((a ∪ b) ∪ d)) = ((a ∪ b) ∪ (((a ∪ b) ∪ c) ∩ d)) | |
| 3 | 1, 2 | ax-r2 36 | . 2 (((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = ((a ∪ b) ∪ (((a ∪ b) ∪ c) ∩ d)) |
| 4 | orcom 73 | . 2 ((a ∪ b) ∪ (((a ∪ b) ∪ c) ∩ d)) = ((((a ∪ b) ∪ c) ∩ d) ∪ (a ∪ b)) | |
| 5 | vneulem6.1 | . . . 4 ((a ∪ b) ∩ (c ∪ d)) = 0 | |
| 6 | 5 | vneulem4 1134 | . . 3 (((a ∪ b) ∪ c) ∩ d) = (c ∩ d) |
| 7 | 6 | ror 71 | . 2 ((((a ∪ b) ∪ c) ∩ d) ∪ (a ∪ b)) = ((c ∩ d) ∪ (a ∪ b)) |
| 8 | 3, 4, 7 | 3tr 65 | 1 (((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = ((c ∩ d) ∪ (a ∪ b)) |
| Colors of variables: term |
| Syntax hints: = wb 1 ∪ wo 6 ∩ wa 7 0wf 9 |
| This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-a3 32 ax-a5 34 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 ax-ml 1122 |
| This theorem depends on definitions: df-a 40 df-t 41 df-f 42 df-le1 130 df-le2 131 |
| This theorem is referenced by: vneulem11 1141 vneulem16 1146 |
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