Metamath Proof Explorer
Description: Equality deduction for a binary relation. (Contributed by NM, 8-Feb-1996) (Proof shortened by Andrew Salmon, 9-Jul-2011)
|
|
Ref |
Expression |
|
Hypotheses |
breq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
|
breq12d.2 |
⊢ ( 𝜑 → 𝐶 = 𝐷 ) |
|
Assertion |
breq12d |
⊢ ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐷 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
breq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
2 |
|
breq12d.2 |
⊢ ( 𝜑 → 𝐶 = 𝐷 ) |
3 |
|
breq12 |
⊢ ( ( 𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ) → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐷 ) ) |
4 |
1 2 3
|
syl2anc |
⊢ ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐷 ) ) |