Proof of Theorem wletr
| Step | Hyp | Ref
| Expression |
| 1 | | wletr.1 |
. . . . . . . 8
(a ≤2 b) = 1 |
| 2 | 1 | wdf-le2 379 |
. . . . . . 7
((a ∪ b) ≡ b) =
1 |
| 3 | 2 | wr5-2v 366 |
. . . . . 6
(((a ∪ b) ∪ c)
≡ (b ∪ c)) = 1 |
| 4 | 3 | wr1 197 |
. . . . 5
((b ∪ c) ≡ ((a
∪ b) ∪ c)) = 1 |
| 5 | | wletr.2 |
. . . . . 6
(b ≤2 c) = 1 |
| 6 | 5 | wdf-le2 379 |
. . . . 5
((b ∪ c) ≡ c) =
1 |
| 7 | | ax-a3 32 |
. . . . . 6
((a ∪ b) ∪ c) =
(a ∪ (b ∪ c)) |
| 8 | 7 | bi1 118 |
. . . . 5
(((a ∪ b) ∪ c)
≡ (a ∪ (b ∪ c))) =
1 |
| 9 | 4, 6, 8 | w3tr2 375 |
. . . 4
(c ≡ (a ∪ (b ∪
c))) = 1 |
| 10 | 9 | wlan 370 |
. . 3
((a ∩ c) ≡ (a
∩ (a ∪ (b ∪ c)))) =
1 |
| 11 | | a5c 121 |
. . . 4
(a ∩ (a ∪ (b ∪
c))) = a |
| 12 | 11 | bi1 118 |
. . 3
((a ∩ (a ∪ (b ∪
c))) ≡ a) = 1 |
| 13 | 10, 12 | wr2 371 |
. 2
((a ∩ c) ≡ a) =
1 |
| 14 | 13 | wdf2le1 385 |
1
(a ≤2 c) = 1 |