Metamath Proof Explorer


Theorem ssnelpss

Description: A subclass missing a member is a proper subclass. (Contributed by NM, 12-Jan-2002)

Ref Expression
Assertion ssnelpss ( 𝐴 ⊆ 𝐵 → ( ( 𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴 ) → 𝐴 ⊊ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 nelneq2 ⊢ ( ( 𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴 ) → ¬ 𝐵 = 𝐴 )
2 1 neqcomd ⊢ ( ( 𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴 ) → ¬ 𝐴 = 𝐵 )
3 dfpss2 ⊢ ( 𝐴 ⊊ 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵 ) )
4 3 baibr ⊢ ( 𝐴 ⊆ 𝐵 → ( ¬ 𝐴 = 𝐵 ↔ 𝐴 ⊊ 𝐵 ) )
5 2 4 imbitrid ⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴 ) → 𝐴 ⊊ 𝐵 ) )