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Mirrors > Home > QLE Home > Th. List > wql1 | GIF version |
Description: The 2nd hypothesis is the first →1 WQL axiom. We show it implies the WOM law. (Contributed by NM, 5-Dec-1998.) |
Ref | Expression |
---|---|
wql1.1 | (a →1 b) = 1 |
wql1.2 | ((a ∪ c) →1 (b ∪ c)) = 1 |
wql1.3 | c = b |
Ref | Expression |
---|---|
wql1 | (a →2 b) = 1 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-i2 45 | . 2 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ )) | |
2 | anor3 90 | . . 3 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥ | |
3 | 2 | lor 70 | . 2 (b ∪ (a⊥ ∩ b⊥ )) = (b ∪ (a ∪ b)⊥ ) |
4 | ax-a2 31 | . . 3 (b ∪ (a ∪ b)⊥ ) = ((a ∪ b)⊥ ∪ b) | |
5 | wql1.3 | . . . . . . . . 9 c = b | |
6 | 5 | lor 70 | . . . . . . . 8 (b ∪ c) = (b ∪ b) |
7 | oridm 110 | . . . . . . . 8 (b ∪ b) = b | |
8 | 6, 7 | ax-r2 36 | . . . . . . 7 (b ∪ c) = b |
9 | 8 | ud1lem0a 255 | . . . . . 6 ((a ∪ c) →1 (b ∪ c)) = ((a ∪ c) →1 b) |
10 | 9 | ax-r1 35 | . . . . 5 ((a ∪ c) →1 b) = ((a ∪ c) →1 (b ∪ c)) |
11 | 5 | lor 70 | . . . . . 6 (a ∪ c) = (a ∪ b) |
12 | 11 | ud1lem0b 256 | . . . . 5 ((a ∪ c) →1 b) = ((a ∪ b) →1 b) |
13 | wql1.2 | . . . . 5 ((a ∪ c) →1 (b ∪ c)) = 1 | |
14 | 10, 12, 13 | 3tr2 64 | . . . 4 ((a ∪ b) →1 b) = 1 |
15 | 14 | wql1lem 287 | . . 3 ((a ∪ b)⊥ ∪ b) = 1 |
16 | 4, 15 | ax-r2 36 | . 2 (b ∪ (a ∪ b)⊥ ) = 1 |
17 | 1, 3, 16 | 3tr 65 | 1 (a →2 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-a4 33 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 df-le1 130 df-le2 131 |
This theorem is referenced by: (None) |
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