Metamath Proof Explorer


Theorem difsnpss

Description: ( B \ { A } ) is a proper subclass of B if and only if A is a member of B . (Contributed by David Moews, 1-May-2017)

Ref Expression
Assertion difsnpss ⊢ A ∈ B ↔ B ∖ A ⊂ B

Proof

Step Hyp Ref Expression
1 notnotb ⊢ A ∈ B ↔ ¬ ¬ A ∈ B
2 difss ⊢ B ∖ A ⊆ B
3 2 biantrur ⊢ B ∖ A ≠ B ↔ B ∖ A ⊆ B ∧ B ∖ A ≠ B
4 difsnb ⊢ ¬ A ∈ B ↔ B ∖ A = B
5 4 necon3bbii ⊢ ¬ ¬ A ∈ B ↔ B ∖ A ≠ B
6 df-pss ⊢ B ∖ A ⊂ B ↔ B ∖ A ⊆ B ∧ B ∖ A ≠ B
7 3 5 6 3bitr4i ⊢ ¬ ¬ A ∈ B ↔ B ∖ A ⊂ B
8 1 7 bitri ⊢ A ∈ B ↔ B ∖ A ⊂ B