Theorem elif0 50013

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.)

Assertion

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

Proof of Theorem elif0

Step	Hyp	Ref	Expression
1		elif 4525	. 2 ⊢ (𝐴 ∈ if(𝜑, 𝐵, ∅) ↔ ((𝜑 ∧ 𝐴 ∈ 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 ∈ ∅)))
2		noel 4283	. . . 4 ⊢ ¬ 𝐴 ∈ ∅
3	2	intnan 492	. . 3 ⊢ ¬ (¬ 𝜑 ∧ 𝐴 ∈ ∅)
4	3	biorfri 953	. 2 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝐵) ↔ ((𝜑 ∧ 𝐴 ∈ 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 ∈ ∅)))
5	1, 4	bitr4i 281	1 ⊢ (𝐴 ∈ if(𝜑, 𝐵, ∅) ↔ (𝜑 ∧ 𝐴 ∈ 𝐵))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401   ∨ wo 861   ∈ wcel 2144  ∅c0 4278  ifcif 4481

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-dif 4047  df-nul 4279  df-if 4482

This theorem is used by:  parhomel  50835  parhomne  50846  parmapcl  50862