Metamath Proof Explorer
Description: Deduction for proof by contradiction. (Contributed by NM, 12-Jun-2014)
|
|
Ref |
Expression |
|
Hypotheses |
pm2.65da.1 |
⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜒 ) |
|
|
pm2.65da.2 |
⊢ ( ( 𝜑 ∧ 𝜓 ) → ¬ 𝜒 ) |
|
Assertion |
pm2.65da |
⊢ ( 𝜑 → ¬ 𝜓 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
pm2.65da.1 |
⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜒 ) |
2 |
|
pm2.65da.2 |
⊢ ( ( 𝜑 ∧ 𝜓 ) → ¬ 𝜒 ) |
3 |
1
|
ex |
⊢ ( 𝜑 → ( 𝜓 → 𝜒 ) ) |
4 |
2
|
ex |
⊢ ( 𝜑 → ( 𝜓 → ¬ 𝜒 ) ) |
5 |
3 4
|
pm2.65d |
⊢ ( 𝜑 → ¬ 𝜓 ) |