Theorem freq2 5618

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 3968	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		frss 5614	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Fr 𝐴 → 𝑅 Fr 𝐵))
3	1, 2	syl 18	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 → 𝑅 Fr 𝐵))
4		eqimss 3967	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		frss 5614	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴))
6	4, 5	syl 18	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴))
7	3, 6	impbid 215	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569   ⊆ wss 3896   Fr wfr 5600

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2752  df-ss 3897  df-fr 5603

This theorem is used by:  freq12d  5619  weeq2  5638  frsn  5739  f1oweALT  7986  frfi  9281  ifr0  45540