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Metamath Proof Explorer

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 ) )