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Proof step 2 from proof diff-del-02 (previous step)

⊦ a ∈ s, s \ s0 ⊆ ((s \ s0) -- a) ++ b

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws)% ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)