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Proof step 4 from proof at-fill-h0 (previous step)

n < # ar, n ≤ m0, ∀ n, ar, a. n < m0 ∧ n < # ar → fill(ar, a, 0, m0)[n] = a ⊦ fill(ar, a, 0, m0)[m0, a][n] = a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ m ≠ n → ar[n, a][m] = ar[m]
Proof step 6:
n < # ar, n ≤ m0, ∀ n, ar, a. n < m0 ∧ n < # ar → fill(ar, a, 0, m0)[n] = a ⊦ fill(ar, a, 0, m0)[n] = a, n = m0

Rule argument

spec: array name: Put-other seq: ⊦ m ≠ n → ar[n, a][m] = ar[m] subst: [m, a, n, ar] → [n, a, m0, fill(ar, a, 0, m0)] flags: <all paths>

Additional information

Used simplifier rules: size-fill-h: ⊦ # fill(ar, a, m, n) = # ar;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%m < # ar ⊦ ar[m, a][m] = a; (s) Used terms: n, a, m0, fill(ar, a, 0, m0)