Theorem eq0rdv

Description: Deduction for equality to the empty set. (Contributed by NM, 11-Jul-2014) Avoid ax-8 , df-clel . (Revised by GG, 6-Sep-2024)

	Ref	Expression
Hypothesis	eq0rdv.1	⊢ ( 𝜑 → ¬ 𝑥 ∈ 𝐴 )
Assertion	eq0rdv	⊢ ( 𝜑 → 𝐴 = ∅ )

Proof

Step	Hyp	Ref	Expression
1		eq0rdv.1	⊢ ( 𝜑 → ¬ 𝑥 ∈ 𝐴 )
2	1	alrimiv	⊢ ( 𝜑 → ∀ 𝑥 ¬ 𝑥 ∈ 𝐴 )
3		biidd	⊢ ( 𝑦 = 𝑥 → ( ⊥ ↔ ⊥ ) )
4	3	eqabbw	⊢ ( 𝐴 = { 𝑦 ∣ ⊥ } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ⊥ ) )
5		dfnul4	⊢ ∅ = { 𝑦 ∣ ⊥ }
6	5	eqeq2i	⊢ ( 𝐴 = ∅ ↔ 𝐴 = { 𝑦 ∣ ⊥ } )
7		nbfal	⊢ ( ¬ 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ↔ ⊥ ) )
8	7	albii	⊢ ( ∀ 𝑥 ¬ 𝑥 ∈ 𝐴 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ⊥ ) )
9	4 6 8	3bitr4ri	⊢ ( ∀ 𝑥 ¬ 𝑥 ∈ 𝐴 ↔ 𝐴 = ∅ )
10	2 9	sylib	⊢ ( 𝜑 → 𝐴 = ∅ )