Metamath Proof Explorer


Theorem mosssn2

Description: Two ways of expressing "at most one" element in a class. (Contributed by Zhi Wang, 23-Sep-2024)

Ref Expression
Assertion mosssn2 ⊢ ∃* x x ∈ A ↔ ∃ y A ⊆ y

Proof

Step Hyp Ref Expression
1 19.45v ⊢ ∃ y A = ∅ ∨ A = y ↔ A = ∅ ∨ ∃ y A = y
2 sssn ⊢ A ⊆ y ↔ A = ∅ ∨ A = y
3 2 exbii ⊢ ∃ y A ⊆ y ↔ ∃ y A = ∅ ∨ A = y
4 mo0sn ⊢ ∃* x x ∈ A ↔ A = ∅ ∨ ∃ y A = y
5 1 3 4 3bitr4ri ⊢ ∃* x x ∈ A ↔ ∃ y A ⊆ y