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Proof step 7 from proof restrict-intersect-empty (previous step)

¬ ¬ a ∈ s, dom st ∩ s = ∅ ⊦ (a ∈ st \ s ↔ a ∈ st) ∧ (st \ s)[a] = st[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
dom st ∩ s = ∅, a ∈ s ⊦ s = ∅, # s = 0, (a ∈ st \ s ↔ a ∈ st) ∧ (∅)[a] = st[a]

Rule argument

Additional information

Used simplifier rules: Diff-mems-get: ⊦ a ∈ s → (st \ s)[a] = (∅)[a];%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic