Theorem freq1 5618

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 5105	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑧𝑅𝑦 ↔ 𝑧𝑆𝑦))
2	1	notbid 321	. . . . 5 ⊢ (𝑅 = 𝑆 → (¬ 𝑧𝑅𝑦 ↔ ¬ 𝑧𝑆𝑦))
3	2	rexralbidv 3229	. . . 4 ⊢ (𝑅 = 𝑆 → (∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦 ↔ ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦))
4	3	imbi2d 343	. . 3 ⊢ (𝑅 = 𝑆 → (((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) ↔ ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦)))
5	4	albidv 1953	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦)))
6		df-fr 5604	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
7		df-fr 5604	. 2 ⊢ (𝑆 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦))
8	5, 6, 7	3bitr4g 317	1 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570   ≠ wne 2956  ∀wral 3077  ∃wrex 3087   ⊆ wss 3899  ∅c0 4279   class class class wbr 5103   Fr wfr 5601

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-br 5104  df-fr 5604

This theorem is used by:  freq12d  5620  weeq1  5638