Metamath Proof Explorer
Description: Cancellation law for subtraction. Deduction form of subcan .
Generalization of subcand . (Contributed by David Moews, 28-Feb-2017)
|
|
Ref |
Expression |
|
Hypotheses |
negidd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
|
|
pncand.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
|
|
subaddd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
|
Assertion |
subcanad |
⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
negidd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
2 |
|
pncand.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
3 |
|
subaddd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
4 |
|
subcan |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) ) |
5 |
1 2 3 4
|
syl3anc |
⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) ) |