Proof of Theorem hfmmvalt
| Step | Hyp | Ref
| Expression |
| 1 | | ax-hilex 8876 |
. . . 4
⊢ ℋ ∈
V |
| 2 | 1 | opabex2 3624 |
. . 3
⊢ {〈x, y〉∣(x ∈ ℋ ⋀ y =
(A · (T ‘x)))}
∈ V |
| 3 | | opreq1 3982 |
. . . . . 6
⊢ (f = A →
(f · (g ‘x)) =
(A · (g ‘x))) |
| 4 | 3 | eqeq2d 1493 |
. . . . 5
⊢ (f = A →
(y = (f
· (g ‘x)) ↔ y =
(A · (g ‘x)))) |
| 5 | 4 | anbi2d 619 |
. . . 4
⊢ (f = A →
((x ∈
ℋ ⋀
y = (f
· (g ‘x))) ↔ (x
∈ ℋ ⋀ y =
(A · (g ‘x))))) |
| 6 | 5 | opabbidv 2683 |
. . 3
⊢ (f = A →
{〈x,
y〉∣(x ∈ ℋ ⋀ y =
(f · (g ‘x)))} =
{〈x,
y〉∣(x ∈ ℋ ⋀ y =
(A · (g ‘x)))}) |
| 7 | | fveq1 3737 |
. . . . . . 7
⊢ (g = T →
(g ‘x) = (T
‘x)) |
| 8 | 7 | opreq2d 3990 |
. . . . . 6
⊢ (g = T →
(A · (g ‘x)) =
(A · (T ‘x))) |
| 9 | 8 | eqeq2d 1493 |
. . . . 5
⊢ (g = T →
(y = (A
· (g ‘x)) ↔ y =
(A · (T ‘x)))) |
| 10 | 9 | anbi2d 619 |
. . . 4
⊢ (g = T →
((x ∈
ℋ ⋀
y = (A
· (g ‘x))) ↔ (x
∈ ℋ ⋀ y =
(A · (T ‘x))))) |
| 11 | 10 | opabbidv 2683 |
. . 3
⊢ (g = T →
{〈x,
y〉∣(x ∈ ℋ ⋀ y =
(A · (g ‘x)))} =
{〈x,
y〉∣(x ∈ ℋ ⋀ y =
(A · (T ‘x)))}) |
| 12 | | df-hfmul 9517 |
. . . 4
⊢
·fn = {〈〈f, g〉, h〉∣((f ∈ ℂ ⋀ g: ℋ –→ℂ) ⋀ h = {〈x, y〉∣(x ∈ ℋ ⋀ y = (f ·
(g ‘x)))})} |
| 13 | | axcnex 5280 |
. . . . . . . 8
⊢ ℂ ∈
V |
| 14 | 13, 1 | elmap 4348 |
. . . . . . 7
⊢ (g ∈ (ℂ ↑m ℋ ) ↔ g:
ℋ –→ℂ) |
| 15 | 14 | anbi2i 483 |
. . . . . 6
⊢ ((f ∈ ℂ ⋀ g ∈ (ℂ ↑m ℋ )) ↔ (f
∈ ℂ ⋀ g: ℋ –→ℂ)) |
| 16 | 15 | anbi1i 484 |
. . . . 5
⊢ (((f ∈ ℂ ⋀ g ∈ (ℂ ↑m ℋ )) ⋀ h = {〈x, y〉∣(x ∈ ℋ ⋀ y = (f ·
(g ‘x)))}) ↔ ((f ∈ ℂ ⋀ g: ℋ
–→ℂ) ⋀ h = {〈x, y〉∣(x ∈ ℋ ⋀ y =
(f · (g ‘x)))})) |
| 17 | 16 | oprabbii 4011 |
. . . 4
⊢ {〈〈f, g〉, h〉∣((f ∈ ℂ ⋀ g ∈ (ℂ ↑m ℋ )) ⋀ h = {〈x, y〉∣(x ∈ ℋ ⋀ y = (f ·
(g ‘x)))})} = {〈〈f, g〉, h〉∣((f ∈ ℂ ⋀ g: ℋ –→ℂ) ⋀ h = {〈x, y〉∣(x ∈ ℋ ⋀ y = (f ·
(g ‘x)))})} |
| 18 | 12, 17 | eqtr4 1505 |
. . 3
⊢
·fn = {〈〈f, g〉, h〉∣((f ∈ ℂ ⋀ g ∈ (ℂ
↑m ℋ )) ⋀ h = {〈x, y〉∣(x ∈ ℋ ⋀ y =
(f · (g ‘x)))})} |
| 19 | 2, 6, 11, 18 | oprabval2 4042 |
. 2
⊢ ((A ∈ ℂ ⋀ T ∈ (ℂ ↑m ℋ )) → (A
·fn T) = {〈x, y〉∣(x ∈ ℋ ⋀ y =
(A · (T ‘x)))}) |
| 20 | 13, 1 | elmap 4348 |
. 2
⊢ (T ∈ (ℂ ↑m ℋ ) ↔ T:
ℋ –→ℂ) |
| 21 | 19, 20 | sylan2br 456 |
1
⊢ ((A ∈ ℂ ⋀ T: ℋ
–→ℂ) → (A ·fn T) = {〈x, y〉∣(x ∈ ℋ ⋀ y = (A ·
(T ‘x)))}) |