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Proof step 24 from proof fill-append-shift (previous step)

n < # ar0 + # ar1, n < # ar0 + # fill(ar1, a, n0, m), # ar0 ≤ n ⊦ # ar1 = 0, (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;ar1[n - # ar0]) = (n < # ar0 ⊃ ar0[n];fill(ar1, a, n0, m)[n - # ar0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
n < # ar0 + # ar1, # ar0 ≤ n ⊦ # ar1 = 0, (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;ar1[n - # ar0]) = fill(ar1, a, n0, m)[n - # ar0]

Rule argument

Additional information

Used simplifier rules: ⊦ # fill(ar, a, m, n) = # ar; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)