Proof of Theorem addcan
| Step | Hyp | Ref
| Expression |
| 1 | | addcan.1 |
. . . 4
⊢ A
∈ ℂ |
| 2 | 1 | cnegex 5329 |
. . 3
⊢ ∃x ∈ ℂ (A + x) =
0 |
| 3 | | addcan.2 |
. . . . . . . . . 10
⊢ B
∈ ℂ |
| 4 | | axaddass 5257 |
. . . . . . . . . 10
⊢ ((x
∈ ℂ ⋀ A ∈ ℂ
⋀ B ∈ ℂ) → ((x + A) +
B) = (x
+ (A + B))) |
| 5 | 3, 4 | mp3an3 903 |
. . . . . . . . 9
⊢ ((x
∈ ℂ ⋀ A ∈ ℂ)
→ ((x + A) + B) =
(x + (A
+ B))) |
| 6 | | addcan.3 |
. . . . . . . . . 10
⊢ C
∈ ℂ |
| 7 | | axaddass 5257 |
. . . . . . . . . 10
⊢ ((x
∈ ℂ ⋀ A ∈ ℂ
⋀ C ∈ ℂ) → ((x + A) +
C) = (x
+ (A + C))) |
| 8 | 6, 7 | mp3an3 903 |
. . . . . . . . 9
⊢ ((x
∈ ℂ ⋀ A ∈ ℂ)
→ ((x + A) + C) =
(x + (A
+ C))) |
| 9 | 5, 8 | eqeq12d 1486 |
. . . . . . . 8
⊢ ((x
∈ ℂ ⋀ A ∈ ℂ)
→ (((x + A) + B) =
((x + A) + C) ↔
(x + (A
+ B)) = (x + (A +
C)))) |
| 10 | 1, 9 | mpan2 695 |
. . . . . . 7
⊢ (x
∈ ℂ → (((x + A) + B) =
((x + A) + C) ↔
(x + (A
+ B)) = (x + (A +
C)))) |
| 11 | | opreq2 3960 |
. . . . . . 7
⊢ ((A +
B) = (A
+ C) → (x + (A +
B)) = (x + (A +
C))) |
| 12 | 10, 11 | syl5bir 210 |
. . . . . 6
⊢ (x
∈ ℂ → ((A + B) = (A +
C) → ((x + A) +
B) = ((x + A) +
C))) |
| 13 | 12 | adantr 389 |
. . . . 5
⊢ ((x
∈ ℂ ⋀ (A + x) = 0) → ((A + B) =
(A + C)
→ ((x + A) + B) =
((x + A) + C))) |
| 14 | | axaddcom 5255 |
. . . . . . . . 9
⊢ ((A
∈ ℂ ⋀ x ∈ ℂ)
→ (A + x) = (x +
A)) |
| 15 | 1, 14 | mpan 694 |
. . . . . . . 8
⊢ (x
∈ ℂ → (A + x) = (x +
A)) |
| 16 | 15 | eqeq1d 1480 |
. . . . . . 7
⊢ (x
∈ ℂ → ((A + x) = 0 ↔ (x
+ A) = 0)) |
| 17 | | opreq1 3959 |
. . . . . . . . 9
⊢ ((x +
A) = 0 → ((x + A) +
B) = (0 + B)) |
| 18 | 3 | addid2 5311 |
. . . . . . . . 9
⊢ (0 + B) = B |
| 19 | 17, 18 | syl6eq 1520 |
. . . . . . . 8
⊢ ((x +
A) = 0 → ((x + A) +
B) = B) |
| 20 | | opreq1 3959 |
. . . . . . . . 9
⊢ ((x +
A) = 0 → ((x + A) +
C) = (0 + C)) |
| 21 | 6 | addid2 5311 |
. . . . . . . . 9
⊢ (0 + C) = C |
| 22 | 20, 21 | syl6eq 1520 |
. . . . . . . 8
⊢ ((x +
A) = 0 → ((x + A) +
C) = C) |
| 23 | 19, 22 | eqeq12d 1486 |
. . . . . . 7
⊢ ((x +
A) = 0 → (((x + A) +
B) = ((x + A) +
C) ↔ B = C)) |
| 24 | 16, 23 | syl6bi 214 |
. . . . . 6
⊢ (x
∈ ℂ → ((A + x) = 0 → (((x + A) +
B) = ((x + A) +
C) ↔ B = C))) |
| 25 | 24 | imp 350 |
. . . . 5
⊢ ((x
∈ ℂ ⋀ (A + x) = 0) → (((x + A) +
B) = ((x + A) +
C) ↔ B = C)) |
| 26 | 13, 25 | sylibd 202 |
. . . 4
⊢ ((x
∈ ℂ ⋀ (A + x) = 0) → ((A + B) =
(A + C)
→ B = C)) |
| 27 | 26 | r19.23aiva 1741 |
. . 3
⊢ (∃x ∈ ℂ (A + x) = 0
→ ((A + B) = (A +
C) → B = C)) |
| 28 | 2, 27 | ax-mp 7 |
. 2
⊢ ((A +
B) = (A
+ C) → B = C) |
| 29 | | opreq2 3960 |
. 2
⊢ (B =
C → (A + B) =
(A + C)) |
| 30 | 28, 29 | impbi 157 |
1
⊢ ((A +
B) = (A
+ C) ↔ B = C) |