Metamath Proof Explorer
Description: EquMoveTerm generator rule. (Contributed by Stanislas Polu, 7-Apr-2020)
|
|
Ref |
Expression |
|
Hypotheses |
int-eqmvtd.1 |
⊢ ( 𝜑 → 𝐶 ∈ ℝ ) |
|
|
int-eqmvtd.2 |
⊢ ( 𝜑 → 𝐷 ∈ ℝ ) |
|
|
int-eqmvtd.3 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
|
int-eqmvtd.4 |
⊢ ( 𝜑 → 𝐴 = ( 𝐶 + 𝐷 ) ) |
|
Assertion |
int-eqmvtd |
⊢ ( 𝜑 → 𝐶 = ( 𝐵 − 𝐷 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
int-eqmvtd.1 |
⊢ ( 𝜑 → 𝐶 ∈ ℝ ) |
| 2 |
|
int-eqmvtd.2 |
⊢ ( 𝜑 → 𝐷 ∈ ℝ ) |
| 3 |
|
int-eqmvtd.3 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
| 4 |
|
int-eqmvtd.4 |
⊢ ( 𝜑 → 𝐴 = ( 𝐶 + 𝐷 ) ) |
| 5 |
3 4
|
eqtr3d |
⊢ ( 𝜑 → 𝐵 = ( 𝐶 + 𝐷 ) ) |
| 6 |
5
|
oveq1d |
⊢ ( 𝜑 → ( 𝐵 − 𝐷 ) = ( ( 𝐶 + 𝐷 ) − 𝐷 ) ) |
| 7 |
1
|
recnd |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
| 8 |
2
|
recnd |
⊢ ( 𝜑 → 𝐷 ∈ ℂ ) |
| 9 |
7 8
|
pncand |
⊢ ( 𝜑 → ( ( 𝐶 + 𝐷 ) − 𝐷 ) = 𝐶 ) |
| 10 |
6 9
|
eqtrd |
⊢ ( 𝜑 → ( 𝐵 − 𝐷 ) = 𝐶 ) |
| 11 |
10
|
eqcomd |
⊢ ( 𝜑 → 𝐶 = ( 𝐵 − 𝐷 ) ) |