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Proof step 3 from proof at-fill-other (previous step)

m < # ar, n < # ar, m ≤ n ⊦ fill(ar, a, 0, m)[n] = ar[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: at-fill-h seq: ⊦ n < # ar → fill(ar, a, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ a;ar[n]) subst: [n, m1, m0, a, ar] → [n, m, 0, a, ar] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: n, m, 0, a, ar