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Proof step 34 from proof from-append-add (previous step)

# x = n2 + n1, n1 < # x, n1 ≤ # x ⊦ x = [], # σ < ⌜n⌝ , #((σ to n) + ⌜x⌝ from n1 + n) = ⌜n2⌝ ∧ (n1 + n0 < # x → ((σ to n) + ⌜x⌝ from n1 + n)[n0] = x[n1 + n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
# x = n2 + n1 ⊦ n2 = 0, x = [], # σ < ⌜n⌝ , #((σ to n) + ⌜x⌝ from n1 + n) = ⌜n2⌝ ∧ (n0 < n2 → ((σ to n) + ⌜x⌝ from n1 + n)[n0] = x[n1 + n0])

Rule argument

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)%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 → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)