Theorem freq2 5668

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 4068	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		frss 5664	. . 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 4067	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		frss 5664	. . 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 212	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1537   ⊆ wss 3976   Fr wfr 5649

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-cleq 2732  df-ss 3993  df-fr 5652

This theorem is referenced by:  freq12d  5669  weeq2  5688  frsn  5787  f1oweALT  8013  frfi  9349  ifr0  44419