Metamath Proof Explorer


Theorem freq12d

Description: Equality deduction for well-founded relations. (Contributed by Stefan O'Rear, 19-Jan-2015) (Proof shortened by Matthew House, 10-Sep-2025)

Ref Expression
Hypotheses freq12d.1 φ R = S
freq12d.2 φ A = B
Assertion freq12d φ R Fr A S Fr B

Proof

Step Hyp Ref Expression
1 freq12d.1 φ R = S
2 freq12d.2 φ A = B
3 freq1 R = S R Fr A S Fr A
4 freq2 A = B S Fr A S Fr B
5 3 4 sylan9bb R = S A = B R Fr A S Fr B
6 1 2 5 syl2anc φ R Fr A S Fr B