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Proof step 4 from proof restrict-intersect-empty (
previous step
)
dom st ∩ s = ∅ ⊦ ∀ a. (a ∈ st \ s ↔ a ∈ st) ∧ (st \ s)[a] = st[a]
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
dom st ∩ s = ∅ ⊦ (a ∈ st \ s ↔ a ∈ st) ∧ (st \ s)[a] = st[a]
Rule argument
Additional information