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Mirrors > Home > QLE Home > Th. List > u3lem11a | GIF version |
Description: Lemma for unified implication study. (Contributed by NM, 18-Jan-1998.) |
Ref | Expression |
---|---|
u3lem11a | (a →3 ((b →3 a) →3 (a →3 b))⊥ ) = (a →3 b⊥ ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | ud3lem1 570 | . . . . 5 ((b →3 a) →3 (a →3 b)) = (b ∪ (b⊥ ∩ a⊥ )) | |
2 | ancom 74 | . . . . . . . 8 (b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ ) | |
3 | anor3 90 | . . . . . . . 8 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥ | |
4 | 2, 3 | ax-r2 36 | . . . . . . 7 (b⊥ ∩ a⊥ ) = (a ∪ b)⊥ |
5 | 4 | lor 70 | . . . . . 6 (b ∪ (b⊥ ∩ a⊥ )) = (b ∪ (a ∪ b)⊥ ) |
6 | oran1 91 | . . . . . 6 (b ∪ (a ∪ b)⊥ ) = (b⊥ ∩ (a ∪ b))⊥ | |
7 | 5, 6 | ax-r2 36 | . . . . 5 (b ∪ (b⊥ ∩ a⊥ )) = (b⊥ ∩ (a ∪ b))⊥ |
8 | 1, 7 | ax-r2 36 | . . . 4 ((b →3 a) →3 (a →3 b)) = (b⊥ ∩ (a ∪ b))⊥ |
9 | 8 | con2 67 | . . 3 ((b →3 a) →3 (a →3 b))⊥ = (b⊥ ∩ (a ∪ b)) |
10 | 9 | ud3lem0a 260 | . 2 (a →3 ((b →3 a) →3 (a →3 b))⊥ ) = (a →3 (b⊥ ∩ (a ∪ b))) |
11 | u3lem11 786 | . 2 (a →3 (b⊥ ∩ (a ∪ b))) = (a →3 b⊥ ) | |
12 | 10, 11 | ax-r2 36 | 1 (a →3 ((b →3 a) →3 (a →3 b))⊥ ) = (a →3 b⊥ ) |
Colors of variables: term |
Syntax hints: = wb 1 ⊥ wn 4 ∪ wo 6 ∩ wa 7 →3 wi3 14 |
This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-a3 32 ax-a4 33 ax-a5 34 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 ax-r3 439 |
This theorem depends on definitions: df-b 39 df-a 40 df-t 41 df-f 42 df-i3 46 df-le1 130 df-le2 131 df-c1 132 df-c2 133 |
This theorem is referenced by: (None) |
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