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Proof step 14 from proof range-copy-drop (previous step)

ar2[n2 + n3] = (n1 + n3 < n0 ∨ n0 + m0 ≤ n1 + n3 ⊃ ar0[n1 + n3];ar[(n1 + n3) + n - n0]), n3 < m1, n1 + n3 < # ar0, n2 + m1 ≤ # ar2, n + m0 ≤ # ar, n0 + m0 ≤ # ar0, n1 + m1 ≤ # ar0, ∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, ar0, n0, m0)[n1 + n3], n0 + m0 ≤ n1 ∨ n1 + m1 ≤ n0 ⊦ ar2[n2 + n3] = ar0[n1 + n3]

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)