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Mirrors > Home > QLE Home > Th. List > 0i1 | GIF version |
Description: Antecedent of 0 on Sasaki conditional. (Contributed by NM, 24-Dec-1998.) |
Ref | Expression |
---|---|
0i1 | (0 →1 a) = 1 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-i1 44 | . 2 (0 →1 a) = (0⊥ ∪ (0 ∩ a)) | |
2 | ax-a2 31 | . . 3 (0⊥ ∪ (0 ∩ a)) = ((0 ∩ a) ∪ 0⊥ ) | |
3 | df-f 42 | . . . . 5 0 = 1⊥ | |
4 | 3 | con2 67 | . . . 4 0⊥ = 1 |
5 | 4 | lor 70 | . . 3 ((0 ∩ a) ∪ 0⊥ ) = ((0 ∩ a) ∪ 1) |
6 | 2, 5 | ax-r2 36 | . 2 (0⊥ ∪ (0 ∩ a)) = ((0 ∩ a) ∪ 1) |
7 | or1 104 | . 2 ((0 ∩ a) ∪ 1) = 1 | |
8 | 1, 6, 7 | 3tr 65 | 1 (0 →1 a) = 1 |
Colors of variables: term |
Syntax hints: = wb 1 ⊥ wn 4 ∪ wo 6 ∩ wa 7 1wt 8 0wf 9 →1 wi1 12 |
This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-a4 33 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 |
This theorem depends on definitions: df-t 41 df-f 42 df-i1 44 |
This theorem is referenced by: oa3-2lema 978 oa3-2to2s 990 |
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