Metamath Proof Explorer


Theorem ssnelpssd

Description: Subclass inclusion with one element of the superclass missing is proper subclass inclusion. Deduction form of ssnelpss . (Contributed by David Moews, 1-May-2017)

Ref Expression
Hypotheses ssnelpssd.1 ⊢ φ → A ⊆ B
ssnelpssd.2 ⊢ φ → C ∈ B
ssnelpssd.3 ⊢ φ → ¬ C ∈ A
Assertion ssnelpssd ⊢ φ → A ⊂ B

Proof

Step Hyp Ref Expression
1 ssnelpssd.1 ⊢ φ → A ⊆ B
2 ssnelpssd.2 ⊢ φ → C ∈ B
3 ssnelpssd.3 ⊢ φ → ¬ C ∈ A
4 ssnelpss ⊢ A ⊆ B → C ∈ B ∧ ¬ C ∈ A → A ⊂ B
5 1 4 syl ⊢ φ → C ∈ B ∧ ¬ C ∈ A → A ⊂ B
6 2 3 5 mp2and ⊢ φ → A ⊂ B