This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem eldifpr

Description: Membership in a set with two elements removed. Similar to eldifsn and eldiftp . (Contributed by Mario Carneiro, 18-Jul-2017)

Ref Expression
Assertion eldifpr ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 , 𝐷 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐴 ≠ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 elprg ⊢ ( 𝐴 ∈ 𝐵 → ( 𝐴 ∈ { 𝐶 , 𝐷 } ↔ ( 𝐴 = 𝐶 ∨ 𝐴 = 𝐷 ) ) )
2 1 notbid ⊢ ( 𝐴 ∈ 𝐵 → ( ¬ 𝐴 ∈ { 𝐶 , 𝐷 } ↔ ¬ ( 𝐴 = 𝐶 ∨ 𝐴 = 𝐷 ) ) )
3 neanior ⊢ ( ( 𝐴 ≠ 𝐶 ∧ 𝐴 ≠ 𝐷 ) ↔ ¬ ( 𝐴 = 𝐶 ∨ 𝐴 = 𝐷 ) )
4 2 3 bitr4di ⊢ ( 𝐴 ∈ 𝐵 → ( ¬ 𝐴 ∈ { 𝐶 , 𝐷 } ↔ ( 𝐴 ≠ 𝐶 ∧ 𝐴 ≠ 𝐷 ) ) )
5 4 pm5.32i ⊢ ( ( 𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ { 𝐶 , 𝐷 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ ( 𝐴 ≠ 𝐶 ∧ 𝐴 ≠ 𝐷 ) ) )
6 eldif ⊢ ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 , 𝐷 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ { 𝐶 , 𝐷 } ) )
7 3anass ⊢ ( ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐴 ≠ 𝐷 ) ↔ ( 𝐴 ∈ 𝐵 ∧ ( 𝐴 ≠ 𝐶 ∧ 𝐴 ≠ 𝐷 ) ) )
8 5 6 7 3bitr4i ⊢ ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 , 𝐷 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ∧ 𝐴 ≠ 𝐷 ) )