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Proof step 13 from proof append-inj (previous step)

# ar = # ar2, # ar0 = # ar1, n < # ar1, n < # ar1 + # ar2 → (ar0 + ar)[n] = (ar1 + ar2)[n], ∀ n. n < # ar1 + # ar2 → (ar0 + ar)[n] = (ar1 + ar2)[n] ⊦ ar0[n] = ar1[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]
Proof step 15:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 16:
# ar0 = # ar1, # ar = # ar2, (ar0 + ar)[n] = ar1[n], n < # ar1 + # ar2, n < # ar1, ∀ n. n < # ar1 + # ar2 → (ar0 + ar)[n] = (ar1 + ar2)[n] ⊦ ar0[n] = ar1[n]

Rule argument

Additional information

Used simplifier rules: append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)