Theorem freq2 5496

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 3950	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		frss 5492	. . 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 3949	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		frss 5492	. . 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 215	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Syntax hints:   → wi 4   ↔ wb 209   = wceq 1539   ⊆ wss 3859   Fr wfr 5481

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-ext 2730

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1542  df-ex 1783  df-sb 2071  df-clab 2737  df-cleq 2751  df-clel 2831  df-v 3412  df-in 3866  df-ss 3876  df-fr 5484

This theorem is referenced by:  weeq2  5514  frsn  5609  f1oweALT  7678  frfi  8789  freq12d  40349  ifr0  41520