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Proof step 13 from proof copy-other-weak (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat