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 )