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

Rule argument

spec: array inst: 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: rewrite

Additional information

Used simplifier rules: size-fill-h: ⊦ # fill(ar, a, m, n) = # ar;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%cgetput: m < # ar ⊦ ar[m, a][m] = a; from specification array Used terms: n, a, m0, fill(ar, a, 0, m0)