Theorem dffr2ALT

Description: Alternate proof of dffr2 , which avoids ax-8 but requires ax-10 , ax-11 , ax-12 . (Contributed by NM, 17-Feb-2004) (Proof shortened by Andrew Salmon, 27-Aug-2011) (Proof shortened by Mario Carneiro, 23-Jun-2015) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	dffr2ALT	⊢ ( 𝑅 Fr 𝐴 ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ) )

Proof

Step	Hyp	Ref	Expression
1		df-fr	⊢ ( 𝑅 Fr 𝐴 ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) )
2		rabeq0	⊢ ( { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ↔ ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 )
3	2	rexbii	⊢ ( ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ↔ ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 )
4	3	imbi2i	⊢ ( ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ) ↔ ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) )
5	4	albii	⊢ ( ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ) ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) )
6	1 5	bitr4i	⊢ ( 𝑅 Fr 𝐴 ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ) )