Metamath Proof Explorer


Theorem limeq

Description: Equality theorem for the limit predicate. (Contributed by NM, 22-Apr-1994) (Proof shortened by Andrew Salmon, 25-Jul-2011)

Ref Expression
Assertion limeq ( 𝐴 = 𝐵 → ( Lim 𝐴 ↔ Lim 𝐵 ) )

Proof

Step Hyp Ref Expression
1 ordeq ⊢ ( 𝐴 = 𝐵 → ( Ord 𝐴 ↔ Ord 𝐵 ) )
2 neeq1 ⊢ ( 𝐴 = 𝐵 → ( 𝐴 ≠ ∅ ↔ 𝐵 ≠ ∅ ) )
3 id ⊢ ( 𝐴 = 𝐵 → 𝐴 = 𝐵 )
4 unieq ⊢ ( 𝐴 = 𝐵 → ∪ 𝐴 = ∪ 𝐵 )
5 3 4 eqeq12d ⊢ ( 𝐴 = 𝐵 → ( 𝐴 = ∪ 𝐴 ↔ 𝐵 = ∪ 𝐵 ) )
6 1 2 5 3anbi123d ⊢ ( 𝐴 = 𝐵 → ( ( Ord 𝐴 ∧ 𝐴 ≠ ∅ ∧ 𝐴 = ∪ 𝐴 ) ↔ ( Ord 𝐵 ∧ 𝐵 ≠ ∅ ∧ 𝐵 = ∪ 𝐵 ) ) )
7 df-lim ⊢ ( Lim 𝐴 ↔ ( Ord 𝐴 ∧ 𝐴 ≠ ∅ ∧ 𝐴 = ∪ 𝐴 ) )
8 df-lim ⊢ ( Lim 𝐵 ↔ ( Ord 𝐵 ∧ 𝐵 ≠ ∅ ∧ 𝐵 = ∪ 𝐵 ) )
9 6 7 8 3bitr4g ⊢ ( 𝐴 = 𝐵 → ( Lim 𝐴 ↔ Lim 𝐵 ) )