Metamath Proof Explorer


Theorem difsnb

Description: ( B \ { A } ) equals B if and only if A is not a member of B . Generalization of difsn . (Contributed by David Moews, 1-May-2017)

Ref Expression
Assertion difsnb ( ¬ 𝐴 ∈ 𝐵 ↔ ( 𝐵 ∖ { 𝐴 } ) = 𝐵 )

Proof

Step Hyp Ref Expression
1 difsn ⊢ ( ¬ 𝐴 ∈ 𝐵 → ( 𝐵 ∖ { 𝐴 } ) = 𝐵 )
2 neldifsnd ⊢ ( 𝐴 ∈ 𝐵 → ¬ 𝐴 ∈ ( 𝐵 ∖ { 𝐴 } ) )
3 nelne1 ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ ( 𝐵 ∖ { 𝐴 } ) ) → 𝐵 ≠ ( 𝐵 ∖ { 𝐴 } ) )
4 2 3 mpdan ⊢ ( 𝐴 ∈ 𝐵 → 𝐵 ≠ ( 𝐵 ∖ { 𝐴 } ) )
5 4 necomd ⊢ ( 𝐴 ∈ 𝐵 → ( 𝐵 ∖ { 𝐴 } ) ≠ 𝐵 )
6 5 necon2bi ⊢ ( ( 𝐵 ∖ { 𝐴 } ) = 𝐵 → ¬ 𝐴 ∈ 𝐵 )
7 1 6 impbii ⊢ ( ¬ 𝐴 ∈ 𝐵 ↔ ( 𝐵 ∖ { 𝐴 } ) = 𝐵 )