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| Mirrors > Home > QLE Home > Th. List > wql2lem4 | GIF version | ||
| Description: Lemma for →2 WQL axiom. (Contributed by NM, 6-Dec-1998.) |
| Ref | Expression |
|---|---|
| wql2lem4.1 | (((a ∩ b⊥ ) ∪ (a ∩ b)) →2 (a⊥ ∪ (a ∩ b))) = 1 |
| wql2lem4.2 | ((a →1 b) ∪ (a ∩ b⊥ )) = 1 |
| Ref | Expression |
|---|---|
| wql2lem4 | (a →1 b) = 1 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | df-i1 44 | . 2 (a →1 b) = (a⊥ ∪ (a ∩ b)) | |
| 2 | id 59 | . 2 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (a ∩ b)) | |
| 3 | ax-a2 31 | . . . 4 ((a ∩ b⊥ ) ∪ (a⊥ ∪ (a ∩ b))) = ((a⊥ ∪ (a ∩ b)) ∪ (a ∩ b⊥ )) | |
| 4 | 1 | ax-r5 38 | . . . . 5 ((a →1 b) ∪ (a ∩ b⊥ )) = ((a⊥ ∪ (a ∩ b)) ∪ (a ∩ b⊥ )) |
| 5 | 4 | ax-r1 35 | . . . 4 ((a⊥ ∪ (a ∩ b)) ∪ (a ∩ b⊥ )) = ((a →1 b) ∪ (a ∩ b⊥ )) |
| 6 | wql2lem4.2 | . . . 4 ((a →1 b) ∪ (a ∩ b⊥ )) = 1 | |
| 7 | 3, 5, 6 | 3tr 65 | . . 3 ((a ∩ b⊥ ) ∪ (a⊥ ∪ (a ∩ b))) = 1 |
| 8 | wql2lem4.1 | . . . 4 (((a ∩ b⊥ ) ∪ (a ∩ b)) →2 (a⊥ ∪ (a ∩ b))) = 1 | |
| 9 | 8 | wql2lem2 289 | . . 3 (((a ∩ b⊥ ) ∪ (a⊥ ∪ (a ∩ b)))⊥ ∪ (a⊥ ∪ (a ∩ b))) = 1 |
| 10 | 7, 9 | skr0 242 | . 2 (a⊥ ∪ (a ∩ b)) = 1 |
| 11 | 1, 2, 10 | 3tr 65 | 1 (a →1 b) = 1 |
| Colors of variables: term |
| Syntax hints: = wb 1 ⊥ wn 4 ∪ wo 6 ∩ wa 7 1wt 8 →1 wi1 12 →2 wi2 13 |
| 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 |
| This theorem depends on definitions: df-a 40 df-t 41 df-f 42 df-i1 44 df-i2 45 |
| This theorem is referenced by: (None) |
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