Theorem freq1 5623

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 9-Mar-1997.)

Assertion

Ref	Expression
freq1	⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Proof of Theorem freq1

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		breq 5127	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑧𝑅𝑦 ↔ 𝑧𝑆𝑦))
2	1	notbid 317	. . . . 5 ⊢ (𝑅 = 𝑆 → (¬ 𝑧𝑅𝑦 ↔ ¬ 𝑧𝑆𝑦))
3	2	rexralbidv 3219	. . . 4 ⊢ (𝑅 = 𝑆 → (∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦 ↔ ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦))
4	3	imbi2d 340	. . 3 ⊢ (𝑅 = 𝑆 → (((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) ↔ ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦)))
5	4	albidv 1923	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦)))
6		df-fr 5608	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
7		df-fr 5608	. 2 ⊢ (𝑆 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦))
8	5, 6, 7	3bitr4g 313	1 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1539   = wceq 1541   ≠ wne 2939  ∀wral 3060  ∃wrex 3069   ⊆ wss 3928  ∅c0 4302   class class class wbr 5125   Fr wfr 5605

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 397  df-ex 1782  df-cleq 2723  df-clel 2809  df-ral 3061  df-rex 3070  df-br 5126  df-fr 5608

This theorem is referenced by:  weeq1  5641  freq12d  41457