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Proof step 2 from proof at-fill-mkarray (previous step)

m < n ⊦ fill(mkarray(n), a, 0, n)[m] = a, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ n < # ar → fill(ar, a, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ a;ar[n])

Rule argument

spec: inst: 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] → [m, n, 0, a, mkarray(n)] flags: rewrite, current

Additional information

Used simplifier rules: lez: ⊦ 0 ≤ m; from specification nat%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%Size-mkarray: ⊦ # mkarray(n) = n; from specification array%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: m, n, 0, a, mkarray(n)