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Proof step 67 from proof splice-resize-less (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 68:
⊦ m < n → resize(ar, n)[m] = (# ar ≤ m ∧ m < n ⊃ zero;ar[m])
Proof step 69:
⊦ # resize(ar, n) = n
Proof step 70:
n < max(# ar1, n1 + m0), n < n2 + n1, # ar1 < n1 + n2, n < n1 + n2, n2 ≤ m0 ⊦ splice(ar0, n0, ar1, n1, m0)[n] = (n < n1 ⊃ (# ar1 ≤ n ⊃ zero;ar1[n]);ar0[n + n0 - n1]), m0 = 0, n2 = 0

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 < n → (n ≤ m ↔ false); (s)%c: ⊦ m + n = n + m; from specification nat-basic