Proof of Theorem ecinxp
Step | Hyp | Ref
| Expression |
1 | | simpr 484 |
. . . . . . . 8
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → 𝐵 ∈ 𝐴) |
2 | 1 | snssd 4834 |
. . . . . . 7
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → {𝐵} ⊆ 𝐴) |
3 | | dfss2 3994 |
. . . . . . 7
⊢ ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∩ 𝐴) = {𝐵}) |
4 | 2, 3 | sylib 218 |
. . . . . 6
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → ({𝐵} ∩ 𝐴) = {𝐵}) |
5 | 4 | imaeq2d 6089 |
. . . . 5
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → (𝑅 “ ({𝐵} ∩ 𝐴)) = (𝑅 “ {𝐵})) |
6 | 5 | ineq1d 4240 |
. . . 4
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → ((𝑅 “ ({𝐵} ∩ 𝐴)) ∩ 𝐴) = ((𝑅 “ {𝐵}) ∩ 𝐴)) |
7 | | imass2 6132 |
. . . . . . 7
⊢ ({𝐵} ⊆ 𝐴 → (𝑅 “ {𝐵}) ⊆ (𝑅 “ 𝐴)) |
8 | 2, 7 | syl 17 |
. . . . . 6
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → (𝑅 “ {𝐵}) ⊆ (𝑅 “ 𝐴)) |
9 | | simpl 482 |
. . . . . 6
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → (𝑅 “ 𝐴) ⊆ 𝐴) |
10 | 8, 9 | sstrd 4019 |
. . . . 5
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → (𝑅 “ {𝐵}) ⊆ 𝐴) |
11 | | dfss2 3994 |
. . . . 5
⊢ ((𝑅 “ {𝐵}) ⊆ 𝐴 ↔ ((𝑅 “ {𝐵}) ∩ 𝐴) = (𝑅 “ {𝐵})) |
12 | 10, 11 | sylib 218 |
. . . 4
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → ((𝑅 “ {𝐵}) ∩ 𝐴) = (𝑅 “ {𝐵})) |
13 | 6, 12 | eqtr2d 2781 |
. . 3
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → (𝑅 “ {𝐵}) = ((𝑅 “ ({𝐵} ∩ 𝐴)) ∩ 𝐴)) |
14 | | imainrect 6212 |
. . 3
⊢ ((𝑅 ∩ (𝐴 × 𝐴)) “ {𝐵}) = ((𝑅 “ ({𝐵} ∩ 𝐴)) ∩ 𝐴) |
15 | 13, 14 | eqtr4di 2798 |
. 2
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → (𝑅 “ {𝐵}) = ((𝑅 ∩ (𝐴 × 𝐴)) “ {𝐵})) |
16 | | df-ec 8765 |
. 2
⊢ [𝐵]𝑅 = (𝑅 “ {𝐵}) |
17 | | df-ec 8765 |
. 2
⊢ [𝐵](𝑅 ∩ (𝐴 × 𝐴)) = ((𝑅 ∩ (𝐴 × 𝐴)) “ {𝐵}) |
18 | 15, 16, 17 | 3eqtr4g 2805 |
1
⊢ (((𝑅 “ 𝐴) ⊆ 𝐴 ∧ 𝐵 ∈ 𝐴) → [𝐵]𝑅 = [𝐵](𝑅 ∩ (𝐴 × 𝐴))) |