Metamath Proof Explorer
Description: Syllogism inference combined with a biconditional. (Contributed by BJ, 25-Apr-2019)
|
|
Ref |
Expression |
|
Hypotheses |
sylanblrc.1 |
⊢ ( 𝜑 → 𝜓 ) |
|
|
sylanblrc.2 |
⊢ 𝜒 |
|
|
sylanblrc.3 |
⊢ ( 𝜃 ↔ ( 𝜓 ∧ 𝜒 ) ) |
|
Assertion |
sylanblrc |
⊢ ( 𝜑 → 𝜃 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sylanblrc.1 |
⊢ ( 𝜑 → 𝜓 ) |
2 |
|
sylanblrc.2 |
⊢ 𝜒 |
3 |
|
sylanblrc.3 |
⊢ ( 𝜃 ↔ ( 𝜓 ∧ 𝜒 ) ) |
4 |
2
|
a1i |
⊢ ( 𝜑 → 𝜒 ) |
5 |
1 4 3
|
sylanbrc |
⊢ ( 𝜑 → 𝜃 ) |