Proof of Theorem stcltr1i
Step | Hyp | Ref
| Expression |
1 | | stcltr1.1 |
. 2
⊢ (𝜑 ↔ (𝑆 ∈ States ∧ ∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(((𝑆‘𝑥) = 1 → (𝑆‘𝑦) = 1) → 𝑥 ⊆ 𝑦))) |
2 | | stcltr1.2 |
. . . 4
⊢ 𝐴 ∈
Cℋ |
3 | | stcltr1.3 |
. . . 4
⊢ 𝐵 ∈
Cℋ |
4 | | fveq2 6103 |
. . . . . . . 8
⊢ (𝑥 = 𝐴 → (𝑆‘𝑥) = (𝑆‘𝐴)) |
5 | 4 | eqeq1d 2612 |
. . . . . . 7
⊢ (𝑥 = 𝐴 → ((𝑆‘𝑥) = 1 ↔ (𝑆‘𝐴) = 1)) |
6 | 5 | imbi1d 330 |
. . . . . 6
⊢ (𝑥 = 𝐴 → (((𝑆‘𝑥) = 1 → (𝑆‘𝑦) = 1) ↔ ((𝑆‘𝐴) = 1 → (𝑆‘𝑦) = 1))) |
7 | | sseq1 3589 |
. . . . . 6
⊢ (𝑥 = 𝐴 → (𝑥 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝑦)) |
8 | 6, 7 | imbi12d 333 |
. . . . 5
⊢ (𝑥 = 𝐴 → ((((𝑆‘𝑥) = 1 → (𝑆‘𝑦) = 1) → 𝑥 ⊆ 𝑦) ↔ (((𝑆‘𝐴) = 1 → (𝑆‘𝑦) = 1) → 𝐴 ⊆ 𝑦))) |
9 | | fveq2 6103 |
. . . . . . . 8
⊢ (𝑦 = 𝐵 → (𝑆‘𝑦) = (𝑆‘𝐵)) |
10 | 9 | eqeq1d 2612 |
. . . . . . 7
⊢ (𝑦 = 𝐵 → ((𝑆‘𝑦) = 1 ↔ (𝑆‘𝐵) = 1)) |
11 | 10 | imbi2d 329 |
. . . . . 6
⊢ (𝑦 = 𝐵 → (((𝑆‘𝐴) = 1 → (𝑆‘𝑦) = 1) ↔ ((𝑆‘𝐴) = 1 → (𝑆‘𝐵) = 1))) |
12 | | sseq2 3590 |
. . . . . 6
⊢ (𝑦 = 𝐵 → (𝐴 ⊆ 𝑦 ↔ 𝐴 ⊆ 𝐵)) |
13 | 11, 12 | imbi12d 333 |
. . . . 5
⊢ (𝑦 = 𝐵 → ((((𝑆‘𝐴) = 1 → (𝑆‘𝑦) = 1) → 𝐴 ⊆ 𝑦) ↔ (((𝑆‘𝐴) = 1 → (𝑆‘𝐵) = 1) → 𝐴 ⊆ 𝐵))) |
14 | 8, 13 | rspc2v 3293 |
. . . 4
⊢ ((𝐴 ∈
Cℋ ∧ 𝐵 ∈ Cℋ )
→ (∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(((𝑆‘𝑥) = 1 → (𝑆‘𝑦) = 1) → 𝑥 ⊆ 𝑦) → (((𝑆‘𝐴) = 1 → (𝑆‘𝐵) = 1) → 𝐴 ⊆ 𝐵))) |
15 | 2, 3, 14 | mp2an 704 |
. . 3
⊢
(∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(((𝑆‘𝑥) = 1 → (𝑆‘𝑦) = 1) → 𝑥 ⊆ 𝑦) → (((𝑆‘𝐴) = 1 → (𝑆‘𝐵) = 1) → 𝐴 ⊆ 𝐵)) |
16 | 15 | adantl 481 |
. 2
⊢ ((𝑆 ∈ States ∧
∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(((𝑆‘𝑥) = 1 → (𝑆‘𝑦) = 1) → 𝑥 ⊆ 𝑦)) → (((𝑆‘𝐴) = 1 → (𝑆‘𝐵) = 1) → 𝐴 ⊆ 𝐵)) |
17 | 1, 16 | sylbi 206 |
1
⊢ (𝜑 → (((𝑆‘𝐴) = 1 → (𝑆‘𝐵) = 1) → 𝐴 ⊆ 𝐵)) |