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Proof step 2 from proof restrict-at (previous step)

a ∈ st, a ∈ s ⊦ (st | s)[a] = st[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ st, a ∈ s ⊦ # s = 0, (st \ (dom st \ s))[a] = st[a], s = ∅

Rule argument

Additional information

Used simplifier rules: restrict: ⊦ st | s = st \ (dom st \ s);%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic