Proof of Theorem stjt
| Step | Hyp | Ref
| Expression |
| 1 | | stelt 5671 |
. . . . 5
⊢ (S
∈ States ↔ ((S:
Cℋ –→ℝ ∧ ∀x ∈ Cℋ (0 ≤
(S ‘x) ∧ (S
‘x) ≤ 1)) ∧ ((S ‘ ℋ ) = 1 ∧ ∀x ∈ Cℋ ∀y ∈ Cℋ (x ⊆ (⊥ ‘y) → (S
‘(x ∨ℋ y)) = ((S
‘x) + (S ‘y)))))) |
| 2 | 1 | pm3.27bd 263 |
. . . 4
⊢ (S
∈ States → ((S ‘ ℋ )
= 1 ∧ ∀x ∈
Cℋ ∀y ∈
Cℋ (x ⊆
(⊥ ‘y) → (S ‘(x
∨ℋ y)) = ((S ‘x) +
(S ‘y))))) |
| 3 | 2 | pm3.27d 262 |
. . 3
⊢ (S
∈ States → ∀x ∈
Cℋ ∀y ∈
Cℋ (x ⊆
(⊥ ‘y) → (S ‘(x
∨ℋ y)) = ((S ‘x) +
(S ‘y)))) |
| 4 | | sseq1 1521 |
. . . . 5
⊢ (x =
A → (x ⊆ (⊥ ‘y) ↔ A
⊆ (⊥ ‘y))) |
| 5 | | opreq1 3006 |
. . . . . . 7
⊢ (x =
A → (x ∨ℋ y) = (A
∨ℋ y)) |
| 6 | 5 | fveq2d 2836 |
. . . . . 6
⊢ (x =
A → (S ‘(x
∨ℋ y)) = (S ‘(A
∨ℋ y))) |
| 7 | | fveq2 2832 |
. . . . . . 7
⊢ (x =
A → (S ‘x) =
(S ‘A)) |
| 8 | 7 | opreq1d 3012 |
. . . . . 6
⊢ (x =
A → ((S ‘x) +
(S ‘y)) = ((S
‘A) + (S ‘y))) |
| 9 | 6, 8 | cleq12d 1115 |
. . . . 5
⊢ (x =
A → ((S ‘(x
∨ℋ y)) = ((S ‘x) +
(S ‘y)) ↔ (S
‘(A ∨ℋ y)) = ((S
‘A) + (S ‘y)))) |
| 10 | 4, 9 | imbi12d 474 |
. . . 4
⊢ (x =
A → ((x ⊆ (⊥ ‘y) → (S
‘(x ∨ℋ y)) = ((S
‘x) + (S ‘y)))
↔ (A ⊆ (⊥ ‘y) → (S
‘(A ∨ℋ y)) = ((S
‘A) + (S ‘y))))) |
| 11 | | fveq2 2832 |
. . . . . 6
⊢ (y =
B → (⊥ ‘y) = (⊥ ‘B)) |
| 12 | 11 | sseq2d 1528 |
. . . . 5
⊢ (y =
B → (A ⊆ (⊥ ‘y) ↔ A
⊆ (⊥ ‘B))) |
| 13 | | opreq2 3007 |
. . . . . . 7
⊢ (y =
B → (A ∨ℋ y) = (A
∨ℋ B)) |
| 14 | 13 | fveq2d 2836 |
. . . . . 6
⊢ (y =
B → (S ‘(A
∨ℋ y)) = (S ‘(A
∨ℋ B))) |
| 15 | | fveq2 2832 |
. . . . . . 7
⊢ (y =
B → (S ‘y) =
(S ‘B)) |
| 16 | 15 | opreq2d 3013 |
. . . . . 6
⊢ (y =
B → ((S ‘A) +
(S ‘y)) = ((S
‘A) + (S ‘B))) |
| 17 | 14, 16 | cleq12d 1115 |
. . . . 5
⊢ (y =
B → ((S ‘(A
∨ℋ y)) = ((S ‘A) +
(S ‘y)) ↔ (S
‘(A ∨ℋ B)) = ((S
‘A) + (S ‘B)))) |
| 18 | 12, 17 | imbi12d 474 |
. . . 4
⊢ (y =
B → ((A ⊆ (⊥ ‘y) → (S
‘(A ∨ℋ y)) = ((S
‘A) + (S ‘y)))
↔ (A ⊆ (⊥ ‘B) → (S
‘(A ∨ℋ B)) = ((S
‘A) + (S ‘B))))) |
| 19 | 10, 18 | rcla42v 1404 |
. . 3
⊢ (∀x ∈ Cℋ ∀y ∈ Cℋ (x ⊆ (⊥ ‘y) → (S
‘(x ∨ℋ y)) = ((S
‘x) + (S ‘y)))
→ ((A ∈
Cℋ ∧ B ∈
Cℋ ) → (A
⊆ (⊥ ‘B) → (S ‘(A
∨ℋ B)) = ((S ‘A) +
(S ‘B))))) |
| 20 | 3, 19 | syl 12 |
. 2
⊢ (S
∈ States → ((A ∈
Cℋ ∧ B ∈
Cℋ ) → (A
⊆ (⊥ ‘B) → (S ‘(A
∨ℋ B)) = ((S ‘A) +
(S ‘B))))) |
| 21 | 20 | imp3a 279 |
1
⊢ (S
∈ States → (((A ∈
Cℋ ∧ B ∈
Cℋ ) ∧ A
⊆ (⊥ ‘B)) →
(S ‘(A ∨ℋ B)) = ((S
‘A) + (S ‘B)))) |