Proof of Theorem subdit
| Step | Hyp | Ref
| Expression |
| 1 | | opreq1 3006 |
. . 3
⊢ (A =
if(A ∈ ℂ, A, 0) → (A
· (B − C)) = (if(A
∈ ℂ, A, 0) · (B − C))) |
| 2 | | opreq1 3006 |
. . . 4
⊢ (A =
if(A ∈ ℂ, A, 0) → (A
· B) = (if(A ∈ ℂ, A, 0) · B)) |
| 3 | | opreq1 3006 |
. . . 4
⊢ (A =
if(A ∈ ℂ, A, 0) → (A
· C) = (if(A ∈ ℂ, A, 0) · C)) |
| 4 | 2, 3 | opreq12d 3014 |
. . 3
⊢ (A =
if(A ∈ ℂ, A, 0) → ((A
· B) − (A · C)) =
((if(A ∈ ℂ, A, 0) · B) − (if(A
∈ ℂ, A, 0) · C))) |
| 5 | 1, 4 | cleq12d 1115 |
. 2
⊢ (A =
if(A ∈ ℂ, A, 0) → ((A
· (B − C)) = ((A
· B) − (A · C))
↔ (if(A ∈ ℂ, A, 0) · (B − C)) =
((if(A ∈ ℂ, A, 0) · B) − (if(A
∈ ℂ, A, 0) · C)))) |
| 6 | | opreq1 3006 |
. . . 4
⊢ (B =
if(B ∈ ℂ, B, 0) → (B
− C) = (if(B ∈ ℂ, B, 0) − C)) |
| 7 | 6 | opreq2d 3013 |
. . 3
⊢ (B =
if(B ∈ ℂ, B, 0) → (if(A ∈ ℂ, A, 0) · (B − C)) =
(if(A ∈ ℂ, A, 0) · (if(B ∈ ℂ, B, 0) − C))) |
| 8 | | opreq2 3007 |
. . . 4
⊢ (B =
if(B ∈ ℂ, B, 0) → (if(A ∈ ℂ, A, 0) · B) = (if(A
∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0))) |
| 9 | 8 | opreq1d 3012 |
. . 3
⊢ (B =
if(B ∈ ℂ, B, 0) → ((if(A ∈ ℂ, A, 0) · B) − (if(A
∈ ℂ, A, 0) · C)) = ((if(A
∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0)) − (if(A ∈ ℂ, A, 0) · C))) |
| 10 | 7, 9 | cleq12d 1115 |
. 2
⊢ (B =
if(B ∈ ℂ, B, 0) → ((if(A ∈ ℂ, A, 0) · (B − C)) =
((if(A ∈ ℂ, A, 0) · B) − (if(A
∈ ℂ, A, 0) · C)) ↔ (if(A
∈ ℂ, A, 0) · (if(B ∈ ℂ, B, 0) − C)) = ((if(A
∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0)) − (if(A ∈ ℂ, A, 0) · C)))) |
| 11 | | opreq2 3007 |
. . . 4
⊢ (C =
if(C ∈ ℂ, C, 0) → (if(B ∈ ℂ, B, 0) − C)
= (if(B ∈ ℂ, B, 0) − if(C ∈ ℂ, C, 0))) |
| 12 | 11 | opreq2d 3013 |
. . 3
⊢ (C =
if(C ∈ ℂ, C, 0) → (if(A ∈ ℂ, A, 0) · (if(B ∈ ℂ, B, 0) − C)) = (if(A
∈ ℂ, A, 0) · (if(B ∈ ℂ, B, 0) − if(C ∈ ℂ, C, 0)))) |
| 13 | | opreq2 3007 |
. . . 4
⊢ (C =
if(C ∈ ℂ, C, 0) → (if(A ∈ ℂ, A, 0) · C) = (if(A
∈ ℂ, A, 0) · if(C ∈ ℂ, C, 0))) |
| 14 | 13 | opreq2d 3013 |
. . 3
⊢ (C =
if(C ∈ ℂ, C, 0) → ((if(A ∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0)) − (if(A ∈ ℂ, A, 0) · C)) = ((if(A
∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0)) − (if(A ∈ ℂ, A, 0) · if(C ∈ ℂ, C, 0)))) |
| 15 | 12, 14 | cleq12d 1115 |
. 2
⊢ (C =
if(C ∈ ℂ, C, 0) → ((if(A ∈ ℂ, A, 0) · (if(B ∈ ℂ, B, 0) − C)) = ((if(A
∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0)) − (if(A ∈ ℂ, A, 0) · C)) ↔ (if(A
∈ ℂ, A, 0) · (if(B ∈ ℂ, B, 0) − if(C ∈ ℂ, C, 0))) = ((if(A
∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0)) − (if(A ∈ ℂ, A, 0) · if(C ∈ ℂ, C, 0))))) |
| 16 | | 0cn 4100 |
. . . 4
⊢ 0 ∈ ℂ |
| 17 | 16 | elimel 1793 |
. . 3
⊢ if(A
∈ ℂ, A, 0) ∈
ℂ |
| 18 | 16 | elimel 1793 |
. . 3
⊢ if(B
∈ ℂ, B, 0) ∈
ℂ |
| 19 | 16 | elimel 1793 |
. . 3
⊢ if(C
∈ ℂ, C, 0) ∈
ℂ |
| 20 | 17, 18, 19 | subdi 4182 |
. 2
⊢ (if(A
∈ ℂ, A, 0) · (if(B ∈ ℂ, B, 0) − if(C ∈ ℂ, C, 0))) = ((if(A
∈ ℂ, A, 0) · if(B ∈ ℂ, B, 0)) − (if(A ∈ ℂ, A, 0) · if(C ∈ ℂ, C, 0))) |
| 21 | 5, 10, 15, 20 | dedth3h 1788 |
1
⊢ ((A
∈ ℂ ∧ B ∈ ℂ ∧
C ∈ ℂ) → (A · (B
− C)) = ((A · B)
− (A · C))) |