Metamath Proof Explorer
Description: Equality deduction for a binary relation. (Contributed by Thierry
Arnoux, 5-Oct-2020)
|
|
Ref |
Expression |
|
Hypotheses |
breq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
|
breqdi.1 |
⊢ ( 𝜑 → 𝐶 𝐴 𝐷 ) |
|
Assertion |
breqdi |
⊢ ( 𝜑 → 𝐶 𝐵 𝐷 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
breq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
2 |
|
breqdi.1 |
⊢ ( 𝜑 → 𝐶 𝐴 𝐷 ) |
3 |
1
|
breqd |
⊢ ( 𝜑 → ( 𝐶 𝐴 𝐷 ↔ 𝐶 𝐵 𝐷 ) ) |
4 |
2 3
|
mpbid |
⊢ ( 𝜑 → 𝐶 𝐵 𝐷 ) |