Theorem freq1 2174

Description: Equality theorem for the founded predicate.

Assertion

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

Proof of Theorem freq1

Step	Hyp	Ref	Expression
1		breq 2064	. . . . . . 7 ⊢ (R = S → (zRy ↔ zSy))
2	1	negbid 463	. . . . . 6 ⊢ (R = S → (¬ zRy ↔ ¬ zSy))
3	2	biraldv 1219	. . . . 5 ⊢ (R = S → (∀z ∈ x ¬ zRy ↔ ∀z ∈ x ¬ zSy))
4	3	birexdv 1220	. . . 4 ⊢ (R = S → (∃y ∈ x ∀z ∈ x ¬ zRy ↔ ∃y ∈ x ∀z ∈ x ¬ zSy))
5	4	imbi2d 464	. . 3 ⊢ (R = S → (((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x ∀z ∈ x ¬ zRy) ↔ ((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x ∀z ∈ x ¬ zSy)))
6	5	bialdv 935	. 2 ⊢ (R = S → (∀x((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x ∀z ∈ x ¬ zRy) ↔ ∀x((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x ∀z ∈ x ¬ zSy)))
7		df-fr 2169	. 2 ⊢ (R Fr A ↔ ∀x((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x ∀z ∈ x ¬ zRy))
8		df-fr 2169	. 2 ⊢ (S Fr A ↔ ∀x((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x ∀z ∈ x ¬ zSy))
9	6, 7, 8	3bitr4g 428	1 ⊢ (R = S → (R Fr A ↔ S Fr A))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672   = wceq 1091  ∀wral 1201  ∃wrex 1202   ⊆ wss 1487  ∅c0 1707   class class class wbr 2054   Fr wfr 2061

This theorem is referenced by:  weeq1 2189

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-br 2063  df-fr 2169

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