Proof of Theorem resoprab
| Step | Hyp | Ref
| Expression |
| 1 | | resopab 3452 |
. . 3
⊢ ({〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ φ)}
↾ (A
× B)) = {〈w, z〉∣(w ∈ (A ×
B) ⋀
∃x∃y(w = 〈x, y〉 ⋀ φ))} |
| 2 | | 19.42vv 1352 |
. . . . 5
⊢ (∃x∃y(w ∈ (A × B)
⋀ (w =
〈x,
y〉 ⋀ φ))
↔ (w ∈ (A ×
B) ⋀
∃x∃y(w = 〈x, y〉 ⋀ φ))) |
| 3 | | an12 495 |
. . . . . . 7
⊢ ((w ∈ (A × B)
⋀ (w =
〈x,
y〉 ⋀ φ))
↔ (w = 〈x, y〉 ⋀ (w ∈ (A ×
B) ⋀
φ))) |
| 4 | | eleq1 1581 |
. . . . . . . . . 10
⊢ (w = 〈x, y〉 → (w
∈ (A
× B) ↔ 〈x, y〉 ∈ (A ×
B))) |
| 5 | | visset 1860 |
. . . . . . . . . . 11
⊢ y ∈
V |
| 6 | 5 | opelxp 3271 |
. . . . . . . . . 10
⊢ (〈x, y〉 ∈ (A ×
B) ↔ (x ∈ A ⋀ y ∈ B)) |
| 7 | 4, 6 | syl6bb 547 |
. . . . . . . . 9
⊢ (w = 〈x, y〉 → (w
∈ (A
× B) ↔ (x ∈ A ⋀ y ∈ B))) |
| 8 | 7 | anbi1d 628 |
. . . . . . . 8
⊢ (w = 〈x, y〉 → ((w
∈ (A
× B) ⋀ φ)
↔ ((x ∈ A ⋀ y ∈ B) ⋀ φ))) |
| 9 | 8 | pm5.32i 656 |
. . . . . . 7
⊢ ((w = 〈x, y〉 ⋀ (w ∈ (A × B)
⋀ φ)) ↔ (w = 〈x, y〉 ⋀ ((x ∈ A ⋀ y ∈ B) ⋀ φ))) |
| 10 | 3, 9 | bitri 180 |
. . . . . 6
⊢ ((w ∈ (A × B)
⋀ (w =
〈x,
y〉 ⋀ φ))
↔ (w = 〈x, y〉 ⋀ ((x ∈ A ⋀ y ∈ B) ⋀ φ))) |
| 11 | 10 | 2exbii 1093 |
. . . . 5
⊢ (∃x∃y(w ∈ (A × B)
⋀ (w =
〈x,
y〉 ⋀ φ))
↔ ∃x∃y(w = 〈x, y〉 ⋀ ((x ∈ A ⋀ y ∈ B) ⋀ φ))) |
| 12 | 2, 11 | bitr3i 182 |
. . . 4
⊢ ((w ∈ (A × B)
⋀ ∃x∃y(w = 〈x, y〉 ⋀ φ)) ↔ ∃x∃y(w = 〈x, y〉 ⋀ ((x ∈ A ⋀ y ∈ B) ⋀ φ))) |
| 13 | 12 | opabbii 2726 |
. . 3
⊢ {〈w, z〉∣(w ∈ (A ×
B) ⋀
∃x∃y(w = 〈x, y〉 ⋀ φ))} = {〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ ((x ∈ A ⋀ y ∈ B) ⋀ φ))} |
| 14 | 1, 13 | eqtri 1542 |
. 2
⊢ ({〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ φ)}
↾ (A
× B)) = {〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ ((x ∈ A ⋀ y ∈ B) ⋀ φ))} |
| 15 | | dfoprab2 4049 |
. . 3
⊢ {〈〈x, y〉, z〉∣φ} = {〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ φ)} |
| 16 | | reseq1 3425 |
. . 3
⊢ ({〈〈x, y〉, z〉∣φ} = {〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ φ)}
→ ({〈〈x, y〉, z〉∣φ} ↾ (A ×
B)) = ({〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ φ)}
↾ (A
× B))) |
| 17 | 15, 16 | ax-mp 7 |
. 2
⊢ ({〈〈x, y〉, z〉∣φ} ↾
(A × B)) = ({〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ φ)} ↾
(A × B)) |
| 18 | | dfoprab2 4049 |
. 2
⊢ {〈〈x, y〉, z〉∣((x ∈ A ⋀ y ∈ B) ⋀ φ)} = {〈w, z〉∣∃x∃y(w = 〈x, y〉 ⋀ ((x ∈ A ⋀ y ∈ B) ⋀ φ))} |
| 19 | 14, 17, 18 | 3eqtr4i 1552 |
1
⊢ ({〈〈x, y〉, z〉∣φ} ↾
(A × B)) = {〈〈x, y〉, z〉∣((x ∈ A ⋀ y ∈ B) ⋀ φ)} |