Metamath Proof Explorer


Theorem elpssex

Description: A class has an element just in case it has a proper subset. (Contributed by Adrian Ducourtial, 26-Oct-2026)

Ref Expression
Assertion elpssex ⊢ ∃ x x ∈ A ↔ ∃ y y ⊂ A

Proof

Step Hyp Ref Expression
1 0ss ⊢ ∅ ⊆ A
2 1 a1i ⊢ x ∈ A → ∅ ⊆ A
3 ne0i ⊢ x ∈ A → A ≠ ∅
4 3 necomd ⊢ x ∈ A → ∅ ≠ A
5 df-pss ⊢ ∅ ⊂ A ↔ ∅ ⊆ A ∧ ∅ ≠ A
6 2 4 5 sylanbrc ⊢ x ∈ A → ∅ ⊂ A
7 0ex ⊢ ∅ ∈ V
8 psseq1 ⊢ y = ∅ → y ⊂ A ↔ ∅ ⊂ A
9 7 8 spcev ⊢ ∅ ⊂ A → ∃ y y ⊂ A
10 6 9 syl ⊢ x ∈ A → ∃ y y ⊂ A
11 10 exlimiv ⊢ ∃ x x ∈ A → ∃ y y ⊂ A
12 pssnel ⊢ y ⊂ A → ∃ x x ∈ A ∧ ¬ x ∈ y
13 exsimpl ⊢ ∃ x x ∈ A ∧ ¬ x ∈ y → ∃ x x ∈ A
14 12 13 syl ⊢ y ⊂ A → ∃ x x ∈ A
15 14 exlimiv ⊢ ∃ y y ⊂ A → ∃ x x ∈ A
16 11 15 impbii ⊢ ∃ x x ∈ A ↔ ∃ y y ⊂ A