Theorem elif0 50127

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 4529	. 2 ⊢ (𝐴 ∈ if(𝜑, 𝐵, ∅) ↔ ((𝜑 ∧ 𝐴 ∈ 𝐵) ∨ (¬ 𝜑 ∧ 𝐴 ∈ ∅)))
2		noel 4286	. . . 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 2147  ∅c0 4281  ifcif 4485

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-dif 4050  df-nul 4282  df-if 4486

This theorem is used by:  parhomel  50949  parhomne  50960  parmapcl  50976