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 e. if ( ph , B , (/) ) <-> ( ph /\ A e. B ) )

Proof

Step Hyp Ref Expression
1 elif
 |-  ( A e. if ( ph , B , (/) ) <-> ( ( ph /\ A e. B ) \/ ( -. ph /\ A e. (/) ) ) )
2 noel
 |-  -. A e. (/)
3 2 intnan
 |-  -. ( -. ph /\ A e. (/) )
4 3 biorfri
 |-  ( ( ph /\ A e. B ) <-> ( ( ph /\ A e. B ) \/ ( -. ph /\ A e. (/) ) ) )
5 1 4 bitr4i
 |-  ( A e. if ( ph , B , (/) ) <-> ( ph /\ A e. B ) )