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Proof step 11 from proof splice-resize-less (previous step)

n2 ≤ m0 ⊦ m0 = 0, # resize(splice(ar0, n0, ar1, n1, m0), n2 + n1) = # resize(splice(ar0, n0, ar1, n1, n2), n2 + n1) ∧ (∀ n. n < # resize(splice(ar0, n0, ar1, n1, m0), n2 + n1) → resize(splice(ar0, n0, ar1, n1, m0), n2 + n1)[n] = resize(splice(ar0, n0, ar1, n1, n2), n2 + n1)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ # resize(ar, n) = n
Proof step 13:
⊦ m < n → resize(ar, n)[m] = (# ar ≤ m ∧ m < n ⊃ zero;ar[m])
Proof step 14:
n < n2 + n1, n2 ≤ m0 ⊦ (# splice(ar0, n0, ar1, n1, m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = (# splice(ar0, n0, ar1, n1, n2) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, n2)[n]), m0 = 0

Rule argument

Additional information

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