Metamath Proof Explorer
Description: Introducing subtraction on both sides of a statement of inequality.
Contrapositive 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 ) |
|
|
subneintrd.4 |
⊢ ( 𝜑 → 𝐵 ≠ 𝐶 ) |
|
Assertion |
subneintrd |
⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) ≠ ( 𝐴 − 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 |
|
subneintrd.4 |
⊢ ( 𝜑 → 𝐵 ≠ 𝐶 ) |
5 |
1 2 3
|
subcanad |
⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) ) |
6 |
5
|
necon3bid |
⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) ≠ ( 𝐴 − 𝐶 ) ↔ 𝐵 ≠ 𝐶 ) ) |
7 |
4 6
|
mpbird |
⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) ≠ ( 𝐴 − 𝐶 ) ) |