Step |
Hyp |
Ref |
Expression |
1 |
|
cnegex |
⊢ ( 𝐴 ∈ ℂ → ∃ 𝑦 ∈ ℂ ( 𝐴 + 𝑦 ) = 0 ) |
2 |
1
|
adantr |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ∃ 𝑦 ∈ ℂ ( 𝐴 + 𝑦 ) = 0 ) |
3 |
|
simpl |
⊢ ( ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) → 𝑦 ∈ ℂ ) |
4 |
|
simpr |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → 𝐵 ∈ ℂ ) |
5 |
|
addcl |
⊢ ( ( 𝑦 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝑦 + 𝐵 ) ∈ ℂ ) |
6 |
3 4 5
|
syl2anr |
⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) → ( 𝑦 + 𝐵 ) ∈ ℂ ) |
7 |
|
simplrr |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( 𝐴 + 𝑦 ) = 0 ) |
8 |
7
|
oveq1d |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( ( 𝐴 + 𝑦 ) + 𝐵 ) = ( 0 + 𝐵 ) ) |
9 |
|
simplll |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → 𝐴 ∈ ℂ ) |
10 |
|
simplrl |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → 𝑦 ∈ ℂ ) |
11 |
|
simpllr |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → 𝐵 ∈ ℂ ) |
12 |
9 10 11
|
addassd |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( ( 𝐴 + 𝑦 ) + 𝐵 ) = ( 𝐴 + ( 𝑦 + 𝐵 ) ) ) |
13 |
11
|
addid2d |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( 0 + 𝐵 ) = 𝐵 ) |
14 |
8 12 13
|
3eqtr3rd |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → 𝐵 = ( 𝐴 + ( 𝑦 + 𝐵 ) ) ) |
15 |
14
|
eqeq2d |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( ( 𝐴 + 𝑥 ) = 𝐵 ↔ ( 𝐴 + 𝑥 ) = ( 𝐴 + ( 𝑦 + 𝐵 ) ) ) ) |
16 |
|
simpr |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → 𝑥 ∈ ℂ ) |
17 |
10 11
|
addcld |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( 𝑦 + 𝐵 ) ∈ ℂ ) |
18 |
9 16 17
|
addcand |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( ( 𝐴 + 𝑥 ) = ( 𝐴 + ( 𝑦 + 𝐵 ) ) ↔ 𝑥 = ( 𝑦 + 𝐵 ) ) ) |
19 |
15 18
|
bitrd |
⊢ ( ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) ∧ 𝑥 ∈ ℂ ) → ( ( 𝐴 + 𝑥 ) = 𝐵 ↔ 𝑥 = ( 𝑦 + 𝐵 ) ) ) |
20 |
19
|
ralrimiva |
⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) → ∀ 𝑥 ∈ ℂ ( ( 𝐴 + 𝑥 ) = 𝐵 ↔ 𝑥 = ( 𝑦 + 𝐵 ) ) ) |
21 |
|
reu6i |
⊢ ( ( ( 𝑦 + 𝐵 ) ∈ ℂ ∧ ∀ 𝑥 ∈ ℂ ( ( 𝐴 + 𝑥 ) = 𝐵 ↔ 𝑥 = ( 𝑦 + 𝐵 ) ) ) → ∃! 𝑥 ∈ ℂ ( 𝐴 + 𝑥 ) = 𝐵 ) |
22 |
6 20 21
|
syl2anc |
⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝑦 ∈ ℂ ∧ ( 𝐴 + 𝑦 ) = 0 ) ) → ∃! 𝑥 ∈ ℂ ( 𝐴 + 𝑥 ) = 𝐵 ) |
23 |
2 22
|
rexlimddv |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ∃! 𝑥 ∈ ℂ ( 𝐴 + 𝑥 ) = 𝐵 ) |