Metamath Proof Explorer
Description: Commutation in consequent. Swap 1st and 2nd. (Contributed by Jeff
Hankins, 17-Nov-2009)
|
|
Ref |
Expression |
|
Hypothesis |
3com12d.1 |
⊢ ( 𝜑 → ( 𝜓 ∧ 𝜒 ∧ 𝜃 ) ) |
|
Assertion |
3com12d |
⊢ ( 𝜑 → ( 𝜒 ∧ 𝜓 ∧ 𝜃 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
3com12d.1 |
⊢ ( 𝜑 → ( 𝜓 ∧ 𝜒 ∧ 𝜃 ) ) |
2 |
|
id |
⊢ ( ( 𝜒 ∧ 𝜓 ∧ 𝜃 ) → ( 𝜒 ∧ 𝜓 ∧ 𝜃 ) ) |
3 |
2
|
3com12 |
⊢ ( ( 𝜓 ∧ 𝜒 ∧ 𝜃 ) → ( 𝜒 ∧ 𝜓 ∧ 𝜃 ) ) |
4 |
1 3
|
syl |
⊢ ( 𝜑 → ( 𝜒 ∧ 𝜓 ∧ 𝜃 ) ) |