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Proof step 22 from proof range-eq-resize (previous step)

n2 < m, m + n1 ≤ # ar1, m + n1 ≤ n, n ≤ # ar1, n0 + m ≤ # ar0, n1 + m ≤ # ar1, ∀ n. n < m → ar0[n0 + n] = ar1[n1 + n] ⊦ ar0[n0 + n2] = resize(ar1, n)[n1 + n2]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ m < n → resize(ar, n)[m] = (# ar ≤ m ∧ m < n ⊃ zero;ar[m])
Proof step 24:
n2 < m, n1 + n2 < n, n1 + m ≤ # ar1, n0 + m ≤ # ar0, n ≤ # ar1, m + n1 ≤ n, m + n1 ≤ # ar1, ∀ n. n < m → ar0[n0 + n] = ar1[n1 + n] ⊦ ar0[n0 + n2] = (# ar1 ≤ n1 + n2 ∧ n1 + n2 < n ⊃ zero;ar1[n1 + n2])

Rule argument

spec: <toplevel> name: resize-at seq: ⊦ m < n → resize(ar, n)[m] = (# ar ≤ m ∧ m < n ⊃ zero;ar[m]) subst: [m, n, ar] → [n1 + n2, n, ar1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1 + n2, n, ar1