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Proof step 11 from proof restrict-intersect-empty (previous step)

a ∈ s, ∀ a. a ∈ dom st ∩ s ↔ a ∈ ∅ ⊦ # s = 0, s = ∅, (a ∈ st \ s ↔ a ∈ st) ∧ (∅)[a] = st[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ s, ∀ a0. ¬ a0 ∈ dom st ∩ s ⊦ # s = 0, s = ∅, (a ∈ st \ s ↔ a ∈ st) ∧ (∅)[a] = st[a]

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)