Step |
Hyp |
Ref |
Expression |
1 |
|
resubval |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 −ℝ 𝐵 ) = ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) ) |
2 |
1
|
eqeq1d |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 𝐴 −ℝ 𝐵 ) = 𝐶 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) ) |
3 |
2
|
3adant3 |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 −ℝ 𝐵 ) = 𝐶 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) ) |
4 |
|
resubeu |
⊢ ( ( 𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ ) → ∃! 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) |
5 |
|
oveq2 |
⊢ ( 𝑥 = 𝐶 → ( 𝐵 + 𝑥 ) = ( 𝐵 + 𝐶 ) ) |
6 |
5
|
eqeq1d |
⊢ ( 𝑥 = 𝐶 → ( ( 𝐵 + 𝑥 ) = 𝐴 ↔ ( 𝐵 + 𝐶 ) = 𝐴 ) ) |
7 |
6
|
riota2 |
⊢ ( ( 𝐶 ∈ ℝ ∧ ∃! 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) ) |
8 |
4 7
|
sylan2 |
⊢ ( ( 𝐶 ∈ ℝ ∧ ( 𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ ) ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) ) |
9 |
8
|
3impb |
⊢ ( ( 𝐶 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) ) |
10 |
9
|
3com13 |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) ) |
11 |
3 10
|
bitr4d |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 −ℝ 𝐵 ) = 𝐶 ↔ ( 𝐵 + 𝐶 ) = 𝐴 ) ) |