Metamath Proof Explorer
Description: Detach truth from conjunction in biconditional. Deduction form.
(Contributed by Peter Mazsa, 24-Sep-2022)
|
|
Ref |
Expression |
|
Hypotheses |
mpbiran2d.1 |
⊢ ( 𝜑 → 𝜃 ) |
|
|
mpbiran2d.2 |
⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) ) |
|
Assertion |
mpbiran2d |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
mpbiran2d.1 |
⊢ ( 𝜑 → 𝜃 ) |
2 |
|
mpbiran2d.2 |
⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) ) |
3 |
2
|
biancomd |
⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜃 ∧ 𝜒 ) ) ) |
4 |
1 3
|
mpbirand |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |