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Proof step 7 from proof append-a (previous step)

n < (# ar + # ar0) + # ar1 ⊦ ((ar + ar0) + ar1)[n] = (ar + ar0 + ar1)[n], n < #(ar + ar0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 9:
n < (# ar + # ar0) + # ar1, # ar + # ar0 ≤ n ⊦ # ar1 = 0, ((ar + ar0) + ar1)[n] = (ar + ar0 + ar1)[n]

Rule argument

Additional information

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