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Proof step 5 from proof append-left-size0 (previous step)

# ar0 = 0 ⊦ #(ar0 + ar1) = # ar1 ∧ (∀ n. n < #(ar0 + ar1) → (ar0 + ar1)[n] = ar1[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 7:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 8:
# ar0 = 0, n < # ar1 ⊦ (ar0 + ar1)[n] = ar1[n]

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n → n ≠ 0; (s)% ⊦ n + 0 = n; (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 = m + n ↔ n = 0; (s)