Step |
Hyp |
Ref |
Expression |
1 |
|
addcom |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐶 ) = ( 𝐶 + 𝐴 ) ) |
2 |
1
|
3adant2 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐶 ) = ( 𝐶 + 𝐴 ) ) |
3 |
|
addcom |
⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) ) |
4 |
3
|
3adant1 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) ) |
5 |
2 4
|
oveq12d |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) ) |
6 |
|
pnpcan |
⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) = ( 𝐴 − 𝐵 ) ) |
7 |
6
|
3coml |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) = ( 𝐴 − 𝐵 ) ) |
8 |
5 7
|
eqtrd |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( 𝐴 − 𝐵 ) ) |