Metamath Proof Explorer
Description: Deduction joining two equivalences to form equivalence of
biconditionals. (Contributed by NM, 26-May-1993)
|
|
Ref |
Expression |
|
Hypotheses |
imbi12d.1 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |
|
|
imbi12d.2 |
⊢ ( 𝜑 → ( 𝜃 ↔ 𝜏 ) ) |
|
Assertion |
bibi12d |
⊢ ( 𝜑 → ( ( 𝜓 ↔ 𝜃 ) ↔ ( 𝜒 ↔ 𝜏 ) ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
imbi12d.1 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |
| 2 |
|
imbi12d.2 |
⊢ ( 𝜑 → ( 𝜃 ↔ 𝜏 ) ) |
| 3 |
1
|
bibi1d |
⊢ ( 𝜑 → ( ( 𝜓 ↔ 𝜃 ) ↔ ( 𝜒 ↔ 𝜃 ) ) ) |
| 4 |
2
|
bibi2d |
⊢ ( 𝜑 → ( ( 𝜒 ↔ 𝜃 ) ↔ ( 𝜒 ↔ 𝜏 ) ) ) |
| 5 |
3 4
|
bitrd |
⊢ ( 𝜑 → ( ( 𝜓 ↔ 𝜃 ) ↔ ( 𝜒 ↔ 𝜏 ) ) ) |