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| Mirrors > Home > QLE Home > Th. List > ud1lem0c | GIF version | ||
| Description: Lemma for unified disjunction. (Contributed by NM, 23-Nov-1997.) |
| Ref | Expression |
|---|---|
| ud1lem0c | (a →1 b)⊥ = (a ∩ (a⊥ ∪ b⊥ )) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | df-i1 44 | . . 3 (a →1 b) = (a⊥ ∪ (a ∩ b)) | |
| 2 | df-a 40 | . . . . . 6 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥ | |
| 3 | df-a 40 | . . . . . . . . 9 (a ∩ b) = (a⊥ ∪ b⊥ )⊥ | |
| 4 | 3 | ax-r1 35 | . . . . . . . 8 (a⊥ ∪ b⊥ )⊥ = (a ∩ b) |
| 5 | 4 | lor 70 | . . . . . . 7 (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ ) = (a⊥ ∪ (a ∩ b)) |
| 6 | 5 | ax-r4 37 | . . . . . 6 (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥ = (a⊥ ∪ (a ∩ b))⊥ |
| 7 | 2, 6 | ax-r2 36 | . . . . 5 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a ∩ b))⊥ |
| 8 | 7 | ax-r1 35 | . . . 4 (a⊥ ∪ (a ∩ b))⊥ = (a ∩ (a⊥ ∪ b⊥ )) |
| 9 | 8 | con3 68 | . . 3 (a⊥ ∪ (a ∩ b)) = (a ∩ (a⊥ ∪ b⊥ ))⊥ |
| 10 | 1, 9 | ax-r2 36 | . 2 (a →1 b) = (a ∩ (a⊥ ∪ b⊥ ))⊥ |
| 11 | 10 | con2 67 | 1 (a →1 b)⊥ = (a ∩ (a⊥ ∪ b⊥ )) |
| Colors of variables: term |
| Syntax hints: = wb 1 ⊥ wn 4 ∪ wo 6 ∩ wa 7 →1 wi1 12 |
| This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 |
| This theorem depends on definitions: df-a 40 df-i1 44 |
| This theorem is referenced by: ud1lem1 560 ud1lem3 562 u1lemc6 706 u1lem11 780 i1abs 801 sa5 836 elimcons2 869 kb10iii 893 |
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