Theorem freq1 5598

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 5087	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑧𝑅𝑦 ↔ 𝑧𝑆𝑦))
2	1	notbid 318	. . . . 5 ⊢ (𝑅 = 𝑆 → (¬ 𝑧𝑅𝑦 ↔ ¬ 𝑧𝑆𝑦))
3	2	rexralbidv 3203	. . . 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 1922	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦)))
6		df-fr 5584	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
7		df-fr 5584	. 2 ⊢ (𝑆 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑆𝑦))
8	5, 6, 7	3bitr4g 314	1 ⊢ (𝑅 = 𝑆 → (𝑅 Fr 𝐴 ↔ 𝑆 Fr 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1540   = wceq 1542   ≠ wne 2932  ∀wral 3051  ∃wrex 3061   ⊆ wss 3889  ∅c0 4273   class class class wbr 5085   Fr wfr 5581

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 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-cleq 2728  df-clel 2811  df-ral 3052  df-rex 3062  df-br 5086  df-fr 5584

This theorem is referenced by:  freq12d  5600  weeq1  5618