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

n < # ar, ∀ n, ar, a, m0. n < # ar → fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ n < m0 + m ⊃ a;ar[n]) ⊦ fill(ar, a, m0, m)[m0 + m, a][n] = (m0 ≤ n ∧ ¬ m0 + m < n ⊃ a;ar[n])

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, ar, a, m0. n < # ar → fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ n < m0 + m ⊃ a;ar[n]) ⊦ fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ ¬ m0 + m < n ⊃ a;ar[n]), n = m0 + m

Rule argument

spec: array name: Put-other seq: ⊦ m ≠ n → ar[n, a][m] = ar[m] subst: [m, a, n, ar] → [n, a, m0 + m, fill(ar, a, m0, m)] 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)% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s) Used terms: n, a, m0 + m, fill(ar, a, m0, m)