Theorem soeq1

Description: Equality theorem for the strict ordering predicate. (Contributed by NM, 16-Mar-1997)

	Ref	Expression
Assertion	soeq1	⊢ ( 𝑅 = 𝑆 → ( 𝑅 Or 𝐴 ↔ 𝑆 Or 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		poeq1	⊢ ( 𝑅 = 𝑆 → ( 𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴 ) )
2		breq	⊢ ( 𝑅 = 𝑆 → ( 𝑥 𝑅 𝑦 ↔ 𝑥 𝑆 𝑦 ) )
3		biidd	⊢ ( 𝑅 = 𝑆 → ( 𝑥 = 𝑦 ↔ 𝑥 = 𝑦 ) )
4		breq	⊢ ( 𝑅 = 𝑆 → ( 𝑦 𝑅 𝑥 ↔ 𝑦 𝑆 𝑥 ) )
5	2 3 4	3orbi123d	⊢ ( 𝑅 = 𝑆 → ( ( 𝑥 𝑅 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑅 𝑥 ) ↔ ( 𝑥 𝑆 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑆 𝑥 ) ) )
6	5	2ralbidv	⊢ ( 𝑅 = 𝑆 → ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ( 𝑥 𝑅 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑅 𝑥 ) ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ( 𝑥 𝑆 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑆 𝑥 ) ) )
7	1 6	anbi12d	⊢ ( 𝑅 = 𝑆 → ( ( 𝑅 Po 𝐴 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ( 𝑥 𝑅 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑅 𝑥 ) ) ↔ ( 𝑆 Po 𝐴 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ( 𝑥 𝑆 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑆 𝑥 ) ) ) )
8		df-so	⊢ ( 𝑅 Or 𝐴 ↔ ( 𝑅 Po 𝐴 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ( 𝑥 𝑅 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑅 𝑥 ) ) )
9		df-so	⊢ ( 𝑆 Or 𝐴 ↔ ( 𝑆 Po 𝐴 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ( 𝑥 𝑆 𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦 𝑆 𝑥 ) ) )
10	7 8 9	3bitr4g	⊢ ( 𝑅 = 𝑆 → ( 𝑅 Or 𝐴 ↔ 𝑆 Or 𝐴 ) )