Metamath Proof Explorer
Description: Join consequents with conjunction. (Contributed by NM, 9-Apr-1994)
|
|
Ref |
Expression |
|
Hypotheses |
3jca.1 |
⊢ ( 𝜑 → 𝜓 ) |
|
|
3jca.2 |
⊢ ( 𝜑 → 𝜒 ) |
|
|
3jca.3 |
⊢ ( 𝜑 → 𝜃 ) |
|
Assertion |
3jca |
⊢ ( 𝜑 → ( 𝜓 ∧ 𝜒 ∧ 𝜃 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
3jca.1 |
⊢ ( 𝜑 → 𝜓 ) |
2 |
|
3jca.2 |
⊢ ( 𝜑 → 𝜒 ) |
3 |
|
3jca.3 |
⊢ ( 𝜑 → 𝜃 ) |
4 |
1 2 3
|
jca31 |
⊢ ( 𝜑 → ( ( 𝜓 ∧ 𝜒 ) ∧ 𝜃 ) ) |
5 |
|
df-3an |
⊢ ( ( 𝜓 ∧ 𝜒 ∧ 𝜃 ) ↔ ( ( 𝜓 ∧ 𝜒 ) ∧ 𝜃 ) ) |
6 |
4 5
|
sylibr |
⊢ ( 𝜑 → ( 𝜓 ∧ 𝜒 ∧ 𝜃 ) ) |