Metamath Proof Explorer
Description: Infer equality from two subclass relationships. Compare Theorem 4 of
Suppes p. 22. (Contributed by NM, 9-Sep-1993)
|
|
Ref |
Expression |
|
Hypotheses |
eqssi.1 |
⊢ 𝐴 ⊆ 𝐵 |
|
|
eqssi.2 |
⊢ 𝐵 ⊆ 𝐴 |
|
Assertion |
eqssi |
⊢ 𝐴 = 𝐵 |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
eqssi.1 |
⊢ 𝐴 ⊆ 𝐵 |
| 2 |
|
eqssi.2 |
⊢ 𝐵 ⊆ 𝐴 |
| 3 |
|
eqss |
⊢ ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) ) |
| 4 |
1 2 3
|
mpbir2an |
⊢ 𝐴 = 𝐵 |