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Proof step 7 from proof at-fill-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m0 < # ar, n < m0 + m, n < # ar, m0 ≤ n, ∀ n, ar, a, m0. n < # ar → fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ n < m0 + m ⊃ a;ar[n]) ∨ # ar = 0 ⊦ m0 + m = 0, # ar = 0, n = m0 + m, fill(ar, a, m0, m)[n] = a, m = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1