Metamath Proof Explorer


Theorem npss

Description: A class is not a proper subclass of another iff it satisfies a one-directional form of eqss . (Contributed by Mario Carneiro, 15-May-2015)

Ref Expression
Assertion npss ( ¬ 𝐴 ⊊ 𝐵 ↔ ( 𝐴 ⊆ 𝐵 → 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 pm4.61 ⊢ ( ¬ ( 𝐴 ⊆ 𝐵 → 𝐴 = 𝐵 ) ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵 ) )
2 dfpss2 ⊢ ( 𝐴 ⊊ 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵 ) )
3 1 2 bitr4i ⊢ ( ¬ ( 𝐴 ⊆ 𝐵 → 𝐴 = 𝐵 ) ↔ 𝐴 ⊊ 𝐵 )
4 3 con1bii ⊢ ( ¬ 𝐴 ⊊ 𝐵 ↔ ( 𝐴 ⊆ 𝐵 → 𝐴 = 𝐵 ) )