Step |
Hyp |
Ref |
Expression |
1 |
|
addcom |
⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) ) |
2 |
1
|
3adant1 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) ) |
3 |
2
|
eqeq1d |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( 𝐶 + 𝐵 ) = 𝐴 ) ) |
4 |
|
subadd |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐵 + 𝐶 ) = 𝐴 ) ) |
5 |
|
subadd |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = 𝐵 ↔ ( 𝐶 + 𝐵 ) = 𝐴 ) ) |
6 |
5
|
3com23 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐶 ) = 𝐵 ↔ ( 𝐶 + 𝐵 ) = 𝐴 ) ) |
7 |
3 4 6
|
3bitr4d |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = 𝐶 ↔ ( 𝐴 − 𝐶 ) = 𝐵 ) ) |