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Proof step 8 from proof owns-none-one2 (conclusion)

⊦ owns-none(a, rs, g) ∧ ¬ r ∈ rs ∧ r ≠ null → owns-one(a, r, rs ++ r, g(r ; a))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : owns-none, , , ⊦ owns-none(a, rs, g) ↔ (∀ r. r ∈ rs → g(r) ≠ a)])]]