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Proof step 2 from proof diff-get-add (previous step)

⊦ (st \ (s ++ a))[b] = st[b], a = b, b ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Diff-rec: ⊦ st \ (s ++ a) = (st \ s) -- a;%Diff-get: ⊦ ¬ a ∈ s → (st \ s)[a] = st[a];% ⊦ a ≠ b → (st -- b)[a] = st[a]; (s)