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Proof step 8 from proof copy-append (previous step)

n0 < # ar1, n0 < n1 + m1 ∨ (n1 + m1) + m0 ≤ n0 ⊦ (n0 < n1 ∨ n1 + m1 ≤ n0 ⊃ ar1[n0];ar0[n0 + n - n1]) = (n0 < n1 ∨ n1 + m0 + m1 ≤ n0 ⊃ ar1[n0];ar0[n0 + n - n1])

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% ⊦ m < n → n ≠ 0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ m + n < m; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)