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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
n < # ar, ∀ n, ar, a, m0. n < # ar → fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ n < m0 + m ⊃ a;ar[n]) ∨ # ar = 0 ⊦ n = m0 + m, # ar = 0, fill(ar, a, m0, m)[m0 + m, a][n] = (m0 ≤ n ∧ ¬ m0 + m < n ⊃ a;ar[n])

Rule argument

fma: n = m0 + m

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%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used formulas: n = m0 + m