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Proof step 13 from proof at-fill-h (previous step)

n < # ar, n < m, # ar ≤ m, ∀ 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

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 14:
n < m, n < # ar, # ar ≤ m, ∀ 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, m0 + m < n

Rule argument

fma: m0 + m < n

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + m0 < n → ¬ n + n0 < m; (s) Used formulas: m0 + m < n