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Proof step 2 from proof diff-add-del-01 (previous step)

⊦ s1 \ (s2 ++ a) = s1 \ s2, a ∈ s1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: diff-add-del: ⊦ s1 \ (s2 ++ a) = (s1 \ s2) -- a;%del-in-03: ⊦ s -- a = s ↔ ¬ a ∈ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;