Metamath Proof Explorer
Description: A contradiction implies anything. Equality/inequality deduction form.
(Contributed by David Moews, 28-Feb-2017)
|
|
Ref |
Expression |
|
Hypotheses |
pm2.21ddne.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
|
pm2.21ddne.2 |
⊢ ( 𝜑 → 𝐴 ≠ 𝐵 ) |
|
Assertion |
pm2.21ddne |
⊢ ( 𝜑 → 𝜓 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
pm2.21ddne.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
2 |
|
pm2.21ddne.2 |
⊢ ( 𝜑 → 𝐴 ≠ 𝐵 ) |
3 |
2
|
neneqd |
⊢ ( 𝜑 → ¬ 𝐴 = 𝐵 ) |
4 |
1 3
|
pm2.21dd |
⊢ ( 𝜑 → 𝜓 ) |