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