Proof of Theorem u3lemax4
| Step | Hyp | Ref
| Expression |
| 1 | | lem4 511 |
. 2
((a →3 b) →3 ((a →3 b) →3 ((b →3 a) →3 ((b →3 a) →3 ((c →3 (c →3 a)) →3 (c →3 (c →3 b))))))) = ((a
→3 b)⊥
∪ ((b →3 a) →3 ((b →3 a) →3 ((c →3 (c →3 a)) →3 (c →3 (c →3 b)))))) |
| 2 | | lem4 511 |
. . . . 5
((b →3 a) →3 ((b →3 a) →3 ((c →3 (c →3 a)) →3 (c →3 (c →3 b))))) = ((b
→3 a)⊥
∪ ((c →3 (c →3 a)) →3 (c →3 (c →3 b)))) |
| 3 | | lem4 511 |
. . . . . . 7
(c →3 (c →3 a)) = (c⊥ ∪ a) |
| 4 | | lem4 511 |
. . . . . . 7
(c →3 (c →3 b)) = (c⊥ ∪ b) |
| 5 | 3, 4 | 2i3 254 |
. . . . . 6
((c →3 (c →3 a)) →3 (c →3 (c →3 b))) = ((c⊥ ∪ a) →3 (c⊥ ∪ b)) |
| 6 | 5 | lor 70 |
. . . . 5
((b →3 a)⊥ ∪ ((c →3 (c →3 a)) →3 (c →3 (c →3 b)))) = ((b
→3 a)⊥
∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b))) |
| 7 | 2, 6 | ax-r2 36 |
. . . 4
((b →3 a) →3 ((b →3 a) →3 ((c →3 (c →3 a)) →3 (c →3 (c →3 b))))) = ((b
→3 a)⊥
∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b))) |
| 8 | 7 | lor 70 |
. . 3
((a →3 b)⊥ ∪ ((b →3 a) →3 ((b →3 a) →3 ((c →3 (c →3 a)) →3 (c →3 (c →3 b)))))) = ((a
→3 b)⊥
∪ ((b →3 a)⊥ ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b)))) |
| 9 | | oran3 93 |
. . . . . 6
((a →3 b)⊥ ∪ (b →3 a)⊥ ) = ((a →3 b) ∩ (b
→3 a))⊥ |
| 10 | | u3lembi 723 |
. . . . . . 7
((a →3 b) ∩ (b
→3 a)) = (a ≡ b) |
| 11 | 10 | ax-r4 37 |
. . . . . 6
((a →3 b) ∩ (b
→3 a))⊥ =
(a ≡ b)⊥ |
| 12 | 9, 11 | ax-r2 36 |
. . . . 5
((a →3 b)⊥ ∪ (b →3 a)⊥ ) = (a ≡ b)⊥ |
| 13 | 12 | ax-r5 38 |
. . . 4
(((a →3 b)⊥ ∪ (b →3 a)⊥ ) ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b))) = ((a
≡ b)⊥ ∪
((c⊥ ∪ a) →3 (c⊥ ∪ b))) |
| 14 | | ax-a3 32 |
. . . 4
(((a →3 b)⊥ ∪ (b →3 a)⊥ ) ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b))) = ((a
→3 b)⊥
∪ ((b →3 a)⊥ ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b)))) |
| 15 | | le1 146 |
. . . . 5
((a ≡ b)⊥ ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b))) ≤ 1 |
| 16 | | ska4 433 |
. . . . . . . 8
((a⊥ ≡
b⊥ )⊥
∪ ((a⊥ ∩ c) ≡ (b⊥ ∩ c))) = 1 |
| 17 | 16 | ax-r1 35 |
. . . . . . 7
1 = ((a⊥ ≡
b⊥ )⊥
∪ ((a⊥ ∩ c) ≡ (b⊥ ∩ c))) |
| 18 | | conb 122 |
. . . . . . . . . 10
(a ≡ b) = (a⊥ ≡ b⊥ ) |
| 19 | 18 | ax-r4 37 |
. . . . . . . . 9
(a ≡ b)⊥ = (a⊥ ≡ b⊥
)⊥ |
| 20 | | conb 122 |
. . . . . . . . . 10
((c⊥ ∪ a) ≡ (c⊥ ∪ b)) = ((c⊥ ∪ a)⊥ ≡ (c⊥ ∪ b)⊥ ) |
| 21 | | ancom 74 |
. . . . . . . . . . . . 13
(a⊥ ∩ c) = (c ∩
a⊥ ) |
| 22 | | anor1 88 |
. . . . . . . . . . . . 13
(c ∩ a⊥ ) = (c⊥ ∪ a)⊥ |
| 23 | 21, 22 | ax-r2 36 |
. . . . . . . . . . . 12
(a⊥ ∩ c) = (c⊥ ∪ a)⊥ |
| 24 | | ancom 74 |
. . . . . . . . . . . . 13
(b⊥ ∩ c) = (c ∩
b⊥ ) |
| 25 | | anor1 88 |
. . . . . . . . . . . . 13
(c ∩ b⊥ ) = (c⊥ ∪ b)⊥ |
| 26 | 24, 25 | ax-r2 36 |
. . . . . . . . . . . 12
(b⊥ ∩ c) = (c⊥ ∪ b)⊥ |
| 27 | 23, 26 | 2bi 99 |
. . . . . . . . . . 11
((a⊥ ∩ c) ≡ (b⊥ ∩ c)) = ((c⊥ ∪ a)⊥ ≡ (c⊥ ∪ b)⊥ ) |
| 28 | 27 | ax-r1 35 |
. . . . . . . . . 10
((c⊥ ∪ a)⊥ ≡ (c⊥ ∪ b)⊥ ) = ((a⊥ ∩ c) ≡ (b⊥ ∩ c)) |
| 29 | 20, 28 | ax-r2 36 |
. . . . . . . . 9
((c⊥ ∪ a) ≡ (c⊥ ∪ b)) = ((a⊥ ∩ c) ≡ (b⊥ ∩ c)) |
| 30 | 19, 29 | 2or 72 |
. . . . . . . 8
((a ≡ b)⊥ ∪ ((c⊥ ∪ a) ≡ (c⊥ ∪ b))) = ((a⊥ ≡ b⊥ )⊥ ∪
((a⊥ ∩ c) ≡ (b⊥ ∩ c))) |
| 31 | 30 | ax-r1 35 |
. . . . . . 7
((a⊥ ≡
b⊥ )⊥
∪ ((a⊥ ∩ c) ≡ (b⊥ ∩ c))) = ((a
≡ b)⊥ ∪
((c⊥ ∪ a) ≡ (c⊥ ∪ b))) |
| 32 | 17, 31 | ax-r2 36 |
. . . . . 6
1 = ((a ≡ b)⊥ ∪ ((c⊥ ∪ a) ≡ (c⊥ ∪ b))) |
| 33 | | u3lembi 723 |
. . . . . . . . 9
(((c⊥ ∪
a) →3 (c⊥ ∪ b)) ∩ ((c⊥ ∪ b) →3 (c⊥ ∪ a))) = ((c⊥ ∪ a) ≡ (c⊥ ∪ b)) |
| 34 | 33 | ax-r1 35 |
. . . . . . . 8
((c⊥ ∪ a) ≡ (c⊥ ∪ b)) = (((c⊥ ∪ a) →3 (c⊥ ∪ b)) ∩ ((c⊥ ∪ b) →3 (c⊥ ∪ a))) |
| 35 | | lea 160 |
. . . . . . . 8
(((c⊥ ∪
a) →3 (c⊥ ∪ b)) ∩ ((c⊥ ∪ b) →3 (c⊥ ∪ a))) ≤ ((c⊥ ∪ a) →3 (c⊥ ∪ b)) |
| 36 | 34, 35 | bltr 138 |
. . . . . . 7
((c⊥ ∪ a) ≡ (c⊥ ∪ b)) ≤ ((c⊥ ∪ a) →3 (c⊥ ∪ b)) |
| 37 | 36 | lelor 166 |
. . . . . 6
((a ≡ b)⊥ ∪ ((c⊥ ∪ a) ≡ (c⊥ ∪ b))) ≤ ((a
≡ b)⊥ ∪
((c⊥ ∪ a) →3 (c⊥ ∪ b))) |
| 38 | 32, 37 | bltr 138 |
. . . . 5
1 ≤ ((a ≡ b)⊥ ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b))) |
| 39 | 15, 38 | lebi 145 |
. . . 4
((a ≡ b)⊥ ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b))) = 1 |
| 40 | 13, 14, 39 | 3tr2 64 |
. . 3
((a →3 b)⊥ ∪ ((b →3 a)⊥ ∪ ((c⊥ ∪ a) →3 (c⊥ ∪ b)))) = 1 |
| 41 | 8, 40 | ax-r2 36 |
. 2
((a →3 b)⊥ ∪ ((b →3 a) →3 ((b →3 a) →3 ((c →3 (c →3 a)) →3 (c →3 (c →3 b)))))) = 1 |
| 42 | 1, 41 | ax-r2 36 |
1
((a →3 b) →3 ((a →3 b) →3 ((b →3 a) →3 ((b →3 a) →3 ((c →3 (c →3 a)) →3 (c →3 (c →3 b))))))) = 1 |