Theorem freq2 5657

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 4055	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		frss 5653	. . 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 4054	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		frss 5653	. . 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 3963   Fr wfr 5638

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1777  df-cleq 2727  df-ss 3980  df-fr 5641

This theorem is referenced by:  freq12d  5658  weeq2  5677  frsn  5776  f1oweALT  7996  frfi  9319  ifr0  44446