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 ) )