Metamath Proof Explorer
Description: Move the right term in a sum on the RHS to the LHS, deduction form.
(Contributed by David A. Wheeler, 11-Oct-2018)
|
|
Ref |
Expression |
|
Hypotheses |
mvrraddd.1 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
|
|
mvrraddd.2 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
|
|
mvrraddd.3 |
⊢ ( 𝜑 → 𝐴 = ( 𝐵 + 𝐶 ) ) |
|
Assertion |
mvrraddd |
⊢ ( 𝜑 → ( 𝐴 − 𝐶 ) = 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
mvrraddd.1 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
2 |
|
mvrraddd.2 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
3 |
|
mvrraddd.3 |
⊢ ( 𝜑 → 𝐴 = ( 𝐵 + 𝐶 ) ) |
4 |
3
|
oveq1d |
⊢ ( 𝜑 → ( 𝐴 − 𝐶 ) = ( ( 𝐵 + 𝐶 ) − 𝐶 ) ) |
5 |
1 2
|
pncand |
⊢ ( 𝜑 → ( ( 𝐵 + 𝐶 ) − 𝐶 ) = 𝐵 ) |
6 |
4 5
|
eqtrd |
⊢ ( 𝜑 → ( 𝐴 − 𝐶 ) = 𝐵 ) |