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Proof step 12 from proof fill-stream-append (previous step)

m + n = # buf, n0 < m + n, n ≤ n0 ⊦ fill(buf, zero, n, m)[n0] = zero, m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ n < # buf → fill(buf, d, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ d;buf[n])

Rule argument

spec: array-copy, inst: buffer name: at-fill-h seq: ⊦ n < # buf → fill(buf, d, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ d;buf[n]) subst: [n, m1, m0, d, buf] → [n0, m, n, zero, buf] flags: <all paths>

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: n0, m, n, zero, buf