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Proof step 11 from proof subarray-copy-higher (previous step)

n < m1, n + m0 < # ar, m0 + m1 ≤ n1, m0 + m1 ≤ # ar ⊦ (n + m0 < n1 ∨ n1 + n2 ≤ n + m0 ⊃ ar[n + m0];ar0[(n + m0) + n0 - n1]) = ar[n + m0]

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 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%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)