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Proof step 18 from proof isempty-ext (previous step)

n0 + n < # buf, isempty(buf, n0, n), ∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty, ∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty ⊦ buf[n0 + n] = empty

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: n0 + n

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s) Used terms: n0 + n