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Proof step 2 from proof at-fill-h (previous step)

n < # ar ⊦ fill(ar, a, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ a;ar[n]), # ar = 0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ n, ar, a, m0. n < # ar → fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ n < m0 + m ⊃ a;ar[n]) ∨ # ar = 0, n < # ar ⊦ fill(ar, a, m0, m +1)[n] = (m0 ≤ n ∧ n < m0 + m +1 ⊃ a;ar[n]), # ar = 0

Rule argument

indvar: m1

Additional information

Used simplifier rules: Fill-zero: ⊦ fill(ar, a, m, 0) = ar;%nat freely generated by 0, +1;;%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat