Proof of Theorem xpundir
| Step | Hyp | Ref
| Expression |
| 1 | | elun 1601 |
. . . . . 6
⊢ (x
∈ (A ∪ B) ↔ (x
∈ A ∨ x ∈ B)) |
| 2 | 1 | anbi1i 368 |
. . . . 5
⊢ ((x
∈ (A ∪ B) ∧ y
∈ C) ↔ ((x ∈ A ∨
x ∈ B) ∧ y
∈ C)) |
| 3 | | andir 457 |
. . . . 5
⊢ (((x
∈ A ∨ x ∈ B)
∧ y ∈ C) ↔ ((x
∈ A ∧ y ∈ C) ∨
(x ∈ B ∧ y ∈
C))) |
| 4 | 2, 3 | bitr 151 |
. . . 4
⊢ ((x
∈ (A ∪ B) ∧ y
∈ C) ↔ ((x ∈ A ∧
y ∈ C) ∨ (x
∈ B ∧ y ∈ C))) |
| 5 | 4 | biopabi 2103 |
. . 3
⊢ {〈x, y〉∣(x
∈ (A ∪ B) ∧ y
∈ C)} = {〈x, y〉∣((x ∈ A ∧
y ∈ C) ∨ (x
∈ B ∧ y ∈ C))} |
| 6 | | unopab 2121 |
. . 3
⊢ ({〈x, y〉∣(x
∈ A ∧ y ∈ C)}
∪ {〈x, y〉∣(x
∈ B ∧ y ∈ C)}) =
{〈x, y〉∣((x ∈ A ∧
y ∈ C) ∨ (x
∈ B ∧ y ∈ C))} |
| 7 | 5, 6 | eqtr4 1122 |
. 2
⊢ {〈x, y〉∣(x
∈ (A ∪ B) ∧ y
∈ C)} = ({〈x, y〉∣(x
∈ A ∧ y ∈ C)}
∪ {〈x, y〉∣(x
∈ B ∧ y ∈ C)}) |
| 8 | | df-xp 2424 |
. 2
⊢ ((A
∪ B) × C) = {〈x,
y〉∣(x ∈ (A
∪ B) ∧ y ∈ C)} |
| 9 | | df-xp 2424 |
. . 3
⊢ (A
× C) = {〈x, y〉∣(x
∈ A ∧ y ∈ C)} |
| 10 | | df-xp 2424 |
. . 3
⊢ (B
× C) = {〈x, y〉∣(x
∈ B ∧ y ∈ C)} |
| 11 | 9, 10 | uneq12i 1609 |
. 2
⊢ ((A
× C) ∪ (B × C)) =
({〈x, y〉∣(x
∈ A ∧ y ∈ C)}
∪ {〈x, y〉∣(x
∈ B ∧ y ∈ C)}) |
| 12 | 7, 8, 11 | 3eqtr4 1126 |
1
⊢ ((A
∪ B) × C) = ((A ×
C) ∪ (B × C)) |