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Proof step 3 from proof append-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 5:
n < (# ar + # ar0) + # ar1 ⊦ ((ar + ar0) + ar1)[n] = (ar + ar0 + ar1)[n]

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # 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 + n0 ↔ n = n0; (s)