Theorem freq1 5610

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 5099	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑧𝑅𝑦 ↔ 𝑧𝑆𝑦))
2	1	notbid 320	. . . . 5 ⊢ (𝑅 = 𝑆 → (¬ 𝑧𝑅𝑦 ↔ ¬ 𝑧𝑆𝑦))
3	2	rexralbidv 3227	. . . 4 ⊢ (𝑅 = 𝑆 → (∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦 ↔ ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦))
4	3	imbi2d 342	. . 3 ⊢ (𝑅 = 𝑆 → (((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) ↔ ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦)))
5	4	albidv 1939	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦)))
6		df-fr 5596	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
7		df-fr 5596	. 2 ⊢ (𝑆 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦))
8	5, 6, 7	3bitr4g 316	1 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 399  ∀wal 1557   = wceq 1559   ≠ wne 2956  ∀wral 3075  ∃wrex 3085   ⊆ wss 3902  ∅c0 4283   class class class wbr 5097   Fr wfr 5593

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1799  df-cleq 2753  df-clel 2836  df-ral 3076  df-rex 3086  df-br 5098  df-fr 5596

This theorem is referenced by:  freq12d  5612  weeq1  5630