Metamath Proof Explorer
Description: Equality implies bijection. (Contributed by RP, 24-Jul-2020)
|
|
Ref |
Expression |
|
Hypothesis |
cleq2lem.b |
⊢ ( 𝐴 = 𝐵 → ( 𝜑 ↔ 𝜓 ) ) |
|
Assertion |
cleq2lem |
⊢ ( 𝐴 = 𝐵 → ( ( 𝑅 ⊆ 𝐴 ∧ 𝜑 ) ↔ ( 𝑅 ⊆ 𝐵 ∧ 𝜓 ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
cleq2lem.b |
⊢ ( 𝐴 = 𝐵 → ( 𝜑 ↔ 𝜓 ) ) |
2 |
|
sseq2 |
⊢ ( 𝐴 = 𝐵 → ( 𝑅 ⊆ 𝐴 ↔ 𝑅 ⊆ 𝐵 ) ) |
3 |
2 1
|
anbi12d |
⊢ ( 𝐴 = 𝐵 → ( ( 𝑅 ⊆ 𝐴 ∧ 𝜑 ) ↔ ( 𝑅 ⊆ 𝐵 ∧ 𝜓 ) ) ) |