¶Intersection of s₁..t₁ and ≠ t₂
The intersection of a range and a non-equivalent depends on how the two types are related.
(s₁..t₁) ∧ (≠ t₂) = ∅ if s₁ = t₁ = t₂
= (s₁..t₁) ∧ (≠ t₂) otherwise
¶Equationally
⟦s₁..t₁⟧ = {x | s₁ ⊆ x ∧ x ⊆ t₁}
⟦≠ t₂⟧ = {x | x ⊈ t₂ ∨ t₂ ⊈ x}
If s₁ = t₁, then the range constraint s₁..t₁ is actually an equivalence constraint = s₁; if also s₁ = t₂, then it's intuitive that the constraints should conflict with each other.
⟦(s..s) ∧ (≠ s)⟧
= {x | s ⊆ x ∧ x ⊆ s} ∩ {x | x ⊈ s ∨ s ⊈ x}
= {x | (s ⊆ x ∧ x ⊆ s) ∧ (x ⊈ s ∨ s ⊈ x)}
= {x | (s ⊆ x ∧ x ⊆ s ∧ x ⊈ s) ∨ (s ⊆ x ∧ x ⊆ s ∧ s ⊈ x)}
= {x | (s ⊆ x ∧ false) ∨ (false ∧ x ⊆ s)}
= {x | false ∨ false}
= {x | false}
= ∅
Otherwise, the two constraints cannot be simplified.