Metamath Proof Explorer
Description: Detach truth from conjunction in biconditional. (Contributed by Glauco
Siliprandi, 3-Mar-2021)
|
|
Ref |
Expression |
|
Hypotheses |
mpbirand.1 |
⊢ ( 𝜑 → 𝜒 ) |
|
|
mpbirand.2 |
⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) ) |
|
Assertion |
mpbirand |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜃 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
mpbirand.1 |
⊢ ( 𝜑 → 𝜒 ) |
2 |
|
mpbirand.2 |
⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) ) |
3 |
1
|
biantrurd |
⊢ ( 𝜑 → ( 𝜃 ↔ ( 𝜒 ∧ 𝜃 ) ) ) |
4 |
2 3
|
bitr4d |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜃 ) ) |