Theorem freq2 2175

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 2173	. . . 4 ⊢ (A ⊆ B → (R Fr B → R Fr A))
2		frss 2173	. . . 4 ⊢ (B ⊆ A → (R Fr A → R Fr B))
3	1, 2	anim12i 268	. . 3 ⊢ ((A ⊆ B ∧ B ⊆ A) → ((R Fr B → R Fr A) ∧ (R Fr A → R Fr B)))
4		eqss 1516	. . 3 ⊢ (A = B ↔ (A ⊆ B ∧ B ⊆ A))
5		bi 396	. . 3 ⊢ ((R Fr B ↔ R Fr A) ↔ ((R Fr B → R Fr A) ∧ (R Fr A → R Fr B)))
6	3, 4, 5	3imtr4 192	. 2 ⊢ (A = B → (R Fr B ↔ R Fr A))
7	6	bicomd 399	1 ⊢ (A = B → (R Fr A ↔ R Fr B))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196   = wceq 1091   ⊆ wss 1487   Fr wfr 2061

This theorem is referenced by:  efrirr 2180  weeq2 2190  f1oweOLD 2944

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-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-fr 2169

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