Metamath Proof Explorer


Theorem ssexnelpss

Description: If there is an element of a class which is not contained in a subclass, the subclass is a proper subclass. (Contributed by AV, 29-Jan-2020)

Ref Expression
Assertion ssexnelpss ⊢ A ⊆ B ∧ ∃ x ∈ B x ∉ A → A ⊂ B

Proof

Step Hyp Ref Expression
1 df-nel ⊢ x ∉ A ↔ ¬ x ∈ A
2 ssnelpss ⊢ A ⊆ B → x ∈ B ∧ ¬ x ∈ A → A ⊂ B
3 2 expdimp ⊢ A ⊆ B ∧ x ∈ B → ¬ x ∈ A → A ⊂ B
4 1 3 biimtrid ⊢ A ⊆ B ∧ x ∈ B → x ∉ A → A ⊂ B
5 4 rexlimdva ⊢ A ⊆ B → ∃ x ∈ B x ∉ A → A ⊂ B
6 5 imp ⊢ A ⊆ B ∧ ∃ x ∈ B x ∉ A → A ⊂ B