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Proof step 1 from proof subarray-append-elim-left0-simp (conclusion)

m ≤ # ar1 ⊦ subarray(ar0 + ar1, # ar0, m) = subarray(ar1, 0, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [subarray(ar1, 0, m), subarray(ar0 + ar1, # ar0, m)] flags: <all paths>

Additional information

Used simplifier rules: append-atp2: n < # ar1 ⊦ (ar0 + ar1)[# ar0 + n] = ar1[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ # subarray(ar, m, n) = n; (s)% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (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%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ n + 0 = n; (s) Used terms: subarray(ar1, 0, m), subarray(ar0 + ar1, # ar0, m)