Theorem freq2 5586

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 3-Apr-1994.)

Assertion

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

Proof of Theorem freq2

Step	Hyp	Ref	Expression
1		eqimss2 3974	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		frss 5582	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Fr 𝐴 → 𝑅 Fr 𝐵))
3	1, 2	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 → 𝑅 Fr 𝐵))
4		eqimss 3973	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		frss 5582	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴))
6	4, 5	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴))
7	3, 6	impbid 213	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Syntax hints:   → wi 4   ↔ wb 207   = wceq 1547   ⊆ wss 3883   Fr wfr 5568

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-cleq 2731  df-ss 3900  df-fr 5571

This theorem is referenced by:  freq12d  5587  weeq2  5606  frsn  5706  f1oweALT  7914  frfi  9185  ifr0  44893