Theorem rabeq0w 4337

Description: Condition for a restricted class abstraction to be empty. Version of rabeq0 4338 using implicit substitution, which does not require ax-10 2178, ax-11 2194, ax-12 2213, but requires ax-8 2147. (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 2844	. . . 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 4328	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = ∅ ↔ ∀𝑦 ¬ (𝑦 ∈ 𝐴 ∧ 𝜓))
5		df-rab 3414	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
6	5	eqeq1i 2766	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = ∅)
7		raln 3086	. 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 1568   = wceq 1570   ∈ wcel 2145  {cab 2739  ∀wral 3077  {crab 3413  ∅c0 4279

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rab 3414  df-dif 3902  df-nul 4280

This theorem is used by:  dffr2  5612  frc  5614  frirr  5627