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Metamath Proof Explorer

Theorem dffr2

Description: Alternate definition of well-founded relation. Similar to Definition 6.21 of TakeutiZaring p. 30. (Contributed by NM, 17-Feb-2004) (Proof shortened by Andrew Salmon, 27-Aug-2011) (Proof shortened by Mario Carneiro, 23-Jun-2015) Avoid ax-10 , ax-11 , ax-12 , but use ax-8 . (Revised by GG, 3-Oct-2024)

Ref Expression
Assertion dffr2 ( 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 ∀ 𝑤 ∈ 𝑥 ¬ 𝑤 𝑅 𝑦 ) )
2 breq1 ⊢ ( 𝑧 = 𝑤 → ( 𝑧 𝑅 𝑦 ↔ 𝑤 𝑅 𝑦 ) )
3 2 rabeq0w ⊢ ( { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ↔ ∀ 𝑤 ∈ 𝑥 ¬ 𝑤 𝑅 𝑦 )
4 3 rexbii ⊢ ( ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ↔ ∃ 𝑦 ∈ 𝑥 ∀ 𝑤 ∈ 𝑥 ¬ 𝑤 𝑅 𝑦 )
5 4 imbi2i ⊢ ( ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ) ↔ ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑤 ∈ 𝑥 ¬ 𝑤 𝑅 𝑦 ) )
6 5 albii ⊢ ( ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ) ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑤 ∈ 𝑥 ¬ 𝑤 𝑅 𝑦 ) )
7 1 6 bitr4i ⊢ ( 𝑅 Fr 𝐴 ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 { 𝑧 ∈ 𝑥 ∣ 𝑧 𝑅 𝑦 } = ∅ ) )