Metamath Proof Explorer


Theorem soeq12d

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

Ref Expression
Hypotheses soeq12d.1 ⊢ φ → R = S
soeq12d.2 ⊢ φ → A = B
Assertion soeq12d ⊢ φ → R Or A ↔ S Or B

Proof

Step Hyp Ref Expression
1 soeq12d.1 ⊢ φ → R = S
2 soeq12d.2 ⊢ φ → A = B
3 soeq1 ⊢ R = S → R Or A ↔ S Or A
4 soeq2 ⊢ A = B → S Or A ↔ S Or B
5 3 4 sylan9bb ⊢ R = S ∧ A = B → R Or A ↔ S Or B
6 1 2 5 syl2anc ⊢ φ → R Or A ↔ S Or B