Metamath Proof Explorer
Description: A modus tollens deduction for inequality. (Contributed by Steven
Nguyen, 1-Jun-2023)
|
|
Ref |
Expression |
|
Hypotheses |
mteqand.1 |
⊢ ( 𝜑 → 𝐶 ≠ 𝐷 ) |
|
|
mteqand.2 |
⊢ ( ( 𝜑 ∧ 𝐴 = 𝐵 ) → 𝐶 = 𝐷 ) |
|
Assertion |
mteqand |
⊢ ( 𝜑 → 𝐴 ≠ 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
mteqand.1 |
⊢ ( 𝜑 → 𝐶 ≠ 𝐷 ) |
2 |
|
mteqand.2 |
⊢ ( ( 𝜑 ∧ 𝐴 = 𝐵 ) → 𝐶 = 𝐷 ) |
3 |
1
|
neneqd |
⊢ ( 𝜑 → ¬ 𝐶 = 𝐷 ) |
4 |
3 2
|
mtand |
⊢ ( 𝜑 → ¬ 𝐴 = 𝐵 ) |
5 |
4
|
neqned |
⊢ ( 𝜑 → 𝐴 ≠ 𝐵 ) |