Theorem freq2 2929

Description: Equality theorem for the founded predicate.

Assertion

Ref	Expression
freq2	⊢ (A = B → (R Fr A ↔ R Fr B))

Proof of Theorem freq2

Step	Hyp	Ref	Expression
1		frss 2927	. . . 4 ⊢ (A ⊆ B → (R Fr B → R Fr A))
2		frss 2927	. . . 4 ⊢ (B ⊆ A → (R Fr A → R Fr B))
3	1, 2	anim12i 333	. . 3 ⊢ ((A ⊆ B ⋀ B ⊆ A) → ((R Fr B → R Fr A) ⋀ (R Fr A → R Fr B)))
4		eqss 2080	. . 3 ⊢ (A = B ↔ (A ⊆ B ⋀ B ⊆ A))
5		dfbi2 516	. . 3 ⊢ ((R Fr B ↔ R Fr A) ↔ ((R Fr B → R Fr A) ⋀ (R Fr A → R Fr B)))
6	3, 4, 5	3imtr4 219	. 2 ⊢ (A = B → (R Fr B ↔ R Fr A))
7	6	bicomd 523	1 ⊢ (A = B → (R Fr A ↔ R Fr B))

Syntax hints:   → wi 3   ↔ wb 146   ⋀ wa 223   = wceq 958   ⊆ wss 2050   Fr wfr 2921

This theorem is referenced by:  efrirr 2934  weeq2 2944  f1oweALT 3912

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-ral 1652  df-rex 1653  df-v 1815  df-dif 2052  df-un 2053  df-in 2054  df-ss 2056  df-nul 2284  df-sn 2416  df-pr 2417  df-op 2420  df-br 2625  df-fr 2923

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