Metamath Proof Explorer


Theorem elif0

Description: Membership in a conditional class with an empty set in the false-branch is equivalent to the condition and membership in the first class. (Contributed by Zhi Wang, 14-Sep-2026)

Ref Expression
Assertion elif0 ( 𝐴 ∈ if ( 𝜑 , 𝐵 , ∅ ) ↔ ( 𝜑 ∧ 𝐴 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 elif ⊢ ( 𝐴 ∈ if ( 𝜑 , 𝐵 , ∅ ) ↔ ( ( 𝜑 ∧ 𝐴 ∈ 𝐵 ) ∨ ( ¬ 𝜑 ∧ 𝐴 ∈ ∅ ) ) )
2 noel ⊢ ¬ 𝐴 ∈ ∅
3 2 intnan ⊢ ¬ ( ¬ 𝜑 ∧ 𝐴 ∈ ∅ )
4 3 biorfri ⊢ ( ( 𝜑 ∧ 𝐴 ∈ 𝐵 ) ↔ ( ( 𝜑 ∧ 𝐴 ∈ 𝐵 ) ∨ ( ¬ 𝜑 ∧ 𝐴 ∈ ∅ ) ) )
5 1 4 bitr4i ⊢ ( 𝐴 ∈ if ( 𝜑 , 𝐵 , ∅ ) ↔ ( 𝜑 ∧ 𝐴 ∈ 𝐵 ) )