Proof of Theorem u5lemob
Step | Hyp | Ref
| Expression |
1 | | df-i5 48 |
. . 3
(a →5 b) = (((a ∩
b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) |
2 | 1 | ax-r5 38 |
. 2
((a →5 b) ∪ b) =
((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ b) |
3 | | ax-a3 32 |
. . 3
((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ b) = (((a ∩
b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∪ b)) |
4 | | lear 161 |
. . . . . 6
(a ∩ b) ≤ b |
5 | | lear 161 |
. . . . . 6
(a⊥ ∩ b) ≤ b |
6 | 4, 5 | lel2or 170 |
. . . . 5
((a ∩ b) ∪ (a⊥ ∩ b)) ≤ b |
7 | | leor 159 |
. . . . 5
b ≤ ((a⊥ ∩ b⊥ ) ∪ b) |
8 | 6, 7 | letr 137 |
. . . 4
((a ∩ b) ∪ (a⊥ ∩ b)) ≤ ((a⊥ ∩ b⊥ ) ∪ b) |
9 | 8 | df-le2 131 |
. . 3
(((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∪ b)) = ((a⊥ ∩ b⊥ ) ∪ b) |
10 | 3, 9 | ax-r2 36 |
. 2
((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b) |
11 | 2, 10 | ax-r2 36 |
1
((a →5 b) ∪ b) =
((a⊥ ∩ b⊥ ) ∪ b) |