Proof of Theorem 4oagen1
| Step | Hyp | Ref
| Expression |
| 1 | | 4oagen1.1 |
. . . . . . 7
g ≤ f |
| 2 | | 4oa.2 |
. . . . . . . 8
f = (((a ∩ b) ∪
((a →1 d) ∩ (b
→1 d))) ∪ e) |
| 3 | | or32 82 |
. . . . . . . 8
(((a ∩ b) ∪ ((a
→1 d) ∩ (b →1 d))) ∪ e) =
(((a ∩ b) ∪ e)
∪ ((a →1 d) ∩ (b
→1 d))) |
| 4 | 2, 3 | ax-r2 36 |
. . . . . . 7
f = (((a ∩ b) ∪
e) ∪ ((a →1 d) ∩ (b
→1 d))) |
| 5 | 1, 4 | lbtr 139 |
. . . . . 6
g ≤ (((a ∩ b) ∪
e) ∪ ((a →1 d) ∩ (b
→1 d))) |
| 6 | 5 | leror 152 |
. . . . 5
(g ∪ ((a →1 d) ∩ (b
→1 d))) ≤ ((((a ∩ b) ∪
e) ∪ ((a →1 d) ∩ (b
→1 d))) ∪ ((a →1 d) ∩ (b
→1 d))) |
| 7 | | ax-a3 32 |
. . . . . 6
((((a ∩ b) ∪ e)
∪ ((a →1 d) ∩ (b
→1 d))) ∪ ((a →1 d) ∩ (b
→1 d))) = (((a ∩ b) ∪
e) ∪ (((a →1 d) ∩ (b
→1 d)) ∪ ((a →1 d) ∩ (b
→1 d)))) |
| 8 | | oridm 110 |
. . . . . . . 8
(((a →1 d) ∩ (b
→1 d)) ∪ ((a →1 d) ∩ (b
→1 d))) = ((a →1 d) ∩ (b
→1 d)) |
| 9 | 8 | lor 70 |
. . . . . . 7
(((a ∩ b) ∪ e)
∪ (((a →1 d) ∩ (b
→1 d)) ∪ ((a →1 d) ∩ (b
→1 d)))) = (((a ∩ b) ∪
e) ∪ ((a →1 d) ∩ (b
→1 d))) |
| 10 | 4 | ax-r1 35 |
. . . . . . 7
(((a ∩ b) ∪ e)
∪ ((a →1 d) ∩ (b
→1 d))) = f |
| 11 | 9, 10 | ax-r2 36 |
. . . . . 6
(((a ∩ b) ∪ e)
∪ (((a →1 d) ∩ (b
→1 d)) ∪ ((a →1 d) ∩ (b
→1 d)))) = f |
| 12 | 7, 11 | ax-r2 36 |
. . . . 5
((((a ∩ b) ∪ e)
∪ ((a →1 d) ∩ (b
→1 d))) ∪ ((a →1 d) ∩ (b
→1 d))) = f |
| 13 | 6, 12 | lbtr 139 |
. . . 4
(g ∪ ((a →1 d) ∩ (b
→1 d))) ≤ f |
| 14 | 13 | lelan 167 |
. . 3
((a →1 d) ∩ (g
∪ ((a →1 d) ∩ (b
→1 d)))) ≤ ((a →1 d) ∩ f) |
| 15 | | 4oa.1 |
. . . 4
e = (((a ∩ c) ∪
((a →1 d) ∩ (c
→1 d))) ∩ ((b ∩ c) ∪
((b →1 d) ∩ (c
→1 d)))) |
| 16 | 15, 2 | 4oath1 1041 |
. . 3
((a →1 d) ∩ f) =
((a →1 d) ∩ (b
→1 d)) |
| 17 | 14, 16 | lbtr 139 |
. 2
((a →1 d) ∩ (g
∪ ((a →1 d) ∩ (b
→1 d)))) ≤ ((a →1 d) ∩ (b
→1 d)) |
| 18 | | lea 160 |
. . 3
((a →1 d) ∩ (b
→1 d)) ≤ (a →1 d) |
| 19 | | leor 159 |
. . 3
((a →1 d) ∩ (b
→1 d)) ≤ (g ∪ ((a
→1 d) ∩ (b →1 d))) |
| 20 | 18, 19 | ler2an 173 |
. 2
((a →1 d) ∩ (b
→1 d)) ≤ ((a →1 d) ∩ (g
∪ ((a →1 d) ∩ (b
→1 d)))) |
| 21 | 17, 20 | lebi 145 |
1
((a →1 d) ∩ (g
∪ ((a →1 d) ∩ (b
→1 d)))) = ((a →1 d) ∩ (b
→1 d)) |