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Proof step 2 from proof below-empty-rem-empty (previous step)

below(ns, n) = ∅ ⊦ below(ns -- m, n) = ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: below-rem-notin: n ≤ m ⊦ below(ns -- m, n) = below(ns, n);%below-rem-in: n > m ⊦ below(ns -- m, n) = below(ns, n) -- m;% ⊦ ∅ -- n = ∅; (s)%size-zero: ⊦ # ns ≠ 0 → ns ≠ ∅; (forward) from specification set-basic[nat-set]% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m > n ↔ n < m; (s)%size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]