Proof of Theorem u4lemonb
Step | Hyp | Ref
| Expression |
1 | | df-i4 47 |
. . 3
(a →4 b) = (((a ∩
b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) |
2 | 1 | ax-r5 38 |
. 2
((a →4 b) ∪ b⊥ ) = ((((a ∩ b) ∪
(a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ ) |
3 | | ax-a3 32 |
. . 3
((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ ) = (((a ∩ b) ∪
(a⊥ ∩ b)) ∪ (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ )) |
4 | | lear 161 |
. . . . 5
((a⊥ ∪ b) ∩ b⊥ ) ≤ b⊥ |
5 | 4 | df-le2 131 |
. . . 4
(((a⊥ ∪
b) ∩ b⊥ ) ∪ b⊥ ) = b⊥ |
6 | 5 | lor 70 |
. . 3
(((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ )) = (((a ∩ b) ∪
(a⊥ ∩ b)) ∪ b⊥ ) |
7 | 3, 6 | ax-r2 36 |
. 2
((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ ) = (((a ∩ b) ∪
(a⊥ ∩ b)) ∪ b⊥ ) |
8 | 2, 7 | ax-r2 36 |
1
((a →4 b) ∪ b⊥ ) = (((a ∩ b) ∪
(a⊥ ∩ b)) ∪ b⊥ ) |