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Proof step 15 from proof rangeeq-emptyleb-datasize (previous step)

# buf = n0 + n1, buf[n0 + (n2 - n0)] = empty, n2 < n0 + n1, n1 ≤ LEB_SIZE, n0 ≤ n2, ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3 ⊦ buf[n2] = empty, n1 = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n2, n0, n4]

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2, n0, n4