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Proof step 28 from proof datasize-empbuf-h (previous step)

m0 < # buf + n, # buf ≤ m0, ∀ buf, n. m0 ≤ # buf + n → datasize(buf + mkempbuf(n), m0) = datasize(buf, min(# buf, m0)) ⊦ m0 + 1 = datasize(buf, # buf), mkempbuf(n)[m0 - # buf] = empty, n = 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] → [m0, # buf, n0]

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)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → mkempbuf(n)[m] = empty; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, # buf, n0