Metamath Proof Explorer


Theorem eldifsn

Description: Membership in a set with an element removed. (Contributed by NM, 10-Oct-2007)

Ref Expression
Assertion eldifsn ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 eldif ⊢ ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ { 𝐶 } ) )
2 elsng ⊢ ( 𝐴 ∈ 𝐵 → ( 𝐴 ∈ { 𝐶 } ↔ 𝐴 = 𝐶 ) )
3 2 necon3bbid ⊢ ( 𝐴 ∈ 𝐵 → ( ¬ 𝐴 ∈ { 𝐶 } ↔ 𝐴 ≠ 𝐶 ) )
4 3 pm5.32i ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ { 𝐶 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ) )
5 1 4 bitri ⊢ ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ) )