Theorem rabeq0w 4336

Description: Condition for a restricted class abstraction to be empty. Version of rabeq0 4337 using implicit substitution, which does not require ax-10 2177, ax-11 2193, ax-12 2212, but requires ax-8 2146. (Contributed by GG, 30-Sep-2024.)

Hypothesis

Ref	Expression
rabeq0w.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
rabeq0w	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} = ∅ ↔ ∀𝑦 ∈ 𝐴 ¬ 𝜓)

Distinct variable groups:   𝑥,𝑦,𝐴   𝜓,𝑥   𝜑,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)

Proof of Theorem rabeq0w

Step	Hyp	Ref	Expression
1		eleq1w 2843	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
2		rabeq0w.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	1, 2	anbi12d 644	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑦 ∈ 𝐴 ∧ 𝜓)))
4	3	ab0w 4327	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = ∅ ↔ ∀𝑦 ¬ (𝑦 ∈ 𝐴 ∧ 𝜓))
5		df-rab 3413	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
6	5	eqeq1i 2765	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = ∅)
7		raln 3085	. 2 ⊢ (∀𝑦 ∈ 𝐴 ¬ 𝜓 ↔ ∀𝑦 ¬ (𝑦 ∈ 𝐴 ∧ 𝜓))
8	4, 6, 7	3bitr4i 306	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} = ∅ ↔ ∀𝑦 ∈ 𝐴 ¬ 𝜓)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567   = wceq 1569   ∈ wcel 2144  {cab 2738  ∀wral 3076  {crab 3412  ∅c0 4278

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-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rab 3413  df-dif 4047  df-nul 4279

This theorem is used by:  dffr2  5611  frc  5613  frirr  5626