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Proof step 56 from proof splice-resize-less (previous step)

max(# ar1, n1 + m0) < n2 + n1, n < n2 + n1, # ar1 < n1 + n2, n < # resize(ar1, n1 + n2), n2 ≤ m0, max(# ar1, n1 + m0) ≤ n ⊦ m0 = 0, n2 = 0, zero = (n < n1 ∨ n1 + n2 ≤ n ⊃ resize(ar1, n1 + n2)[n];ar0[n + n0 - n1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
⊦ m < n → resize(ar, n)[m] = (# ar ≤ m ∧ m < n ⊃ zero;ar[m])
Proof step 58:
⊦ # resize(ar, n) = n
Proof step 59:
# ar1 < n1 + n2, n < n2 + n1, max(# ar1, n1 + m0) < n2 + n1, n < n1 + n2, n1 ≤ n, max(# ar1, n1 + m0) ≤ n, n2 ≤ m0 ⊦ m0 = 0, n2 = 0, zero = ar0[(n - n1) + n0]

Rule argument

Additional information

Used simplifier rules: resize-at: ⊦ m < n → resize(ar, n)[m] = (# ar ≤ m ∧ m < n ⊃ zero;ar[m]);%resize-len: ⊦ # resize(ar, n) = n;% ⊦ m < m + n ↔ n ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ¬ max(n, m) < n; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%c: ⊦ m + n = n + m; from specification nat-basic