Proof of Theorem fnfvbr
| Step | Hyp | Ref
| Expression |
| 1 | | fnfvbr.1 |
. 2
⊢ C
∈ V |
| 2 | | cleq2 1110 |
. . . 4
⊢ (x =
C → ((F ‘B) =
x ↔ (F ‘B) =
C)) |
| 3 | | breq2 2066 |
. . . 4
⊢ (x =
C → (BFx ↔ BFC)) |
| 4 | 2, 3 | bibi12d 477 |
. . 3
⊢ (x =
C → (((F ‘B) =
x ↔ BFx) ↔ ((F
‘B) = C ↔ BFC))) |
| 5 | 4 | imbi2d 464 |
. 2
⊢ (x =
C → (((F Fn A ∧
B ∈ A) → ((F
‘B) = x ↔ BFx)) ↔ ((F
Fn A ∧ B ∈ A)
→ ((F ‘B) = C ↔
BFC)))) |
| 6 | | fneu 2728 |
. . 3
⊢ ((F Fn
A ∧ B ∈ A)
→ ∃!x BFx) |
| 7 | | breq1 2065 |
. . . . . . 7
⊢ (y =
B → (yFx ↔ BFx)) |
| 8 | 7 | bieudv 1013 |
. . . . . 6
⊢ (y =
B → (∃!x yFx ↔
∃!x BFx)) |
| 9 | | fveq2 2832 |
. . . . . . . 8
⊢ (y =
B → (F ‘y) =
(F ‘B)) |
| 10 | 9 | cleq1d 1109 |
. . . . . . 7
⊢ (y =
B → ((F ‘y) =
x ↔ (F ‘B) =
x)) |
| 11 | 10, 7 | bibi12d 477 |
. . . . . 6
⊢ (y =
B → (((F ‘y) =
x ↔ yFx) ↔ ((F
‘B) = x ↔ BFx))) |
| 12 | 8, 11 | imbi12d 474 |
. . . . 5
⊢ (y =
B → ((∃!x yFx →
((F ‘y) = x ↔
yFx)) ↔
(∃!x BFx → ((F
‘B) = x ↔ BFx)))) |
| 13 | | visset 1350 |
. . . . . 6
⊢ y
∈ V |
| 14 | 13 | tz6.12c 2846 |
. . . . 5
⊢ (∃!x yFx →
((F ‘y) = x ↔
yFx)) |
| 15 | 12, 14 | vtoclg 1383 |
. . . 4
⊢ (B
∈ A → (∃!x BFx →
((F ‘B) = x ↔
BFx))) |
| 16 | 15 | adantl 305 |
. . 3
⊢ ((F Fn
A ∧ B ∈ A)
→ (∃!x BFx → ((F
‘B) = x ↔ BFx))) |
| 17 | 6, 16 | mpd 46 |
. 2
⊢ ((F Fn
A ∧ B ∈ A)
→ ((F ‘B) = x ↔
BFx)) |
| 18 | 1, 5, 17 | vtocl 1378 |
1
⊢ ((F Fn
A ∧ B ∈ A)
→ ((F ‘B) = C ↔
BFC)) |