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Proof step 33 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 34:
# x = n2 + n1 ⊦ n2 = 0, x = [], n2 + n1 = 0, # 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: l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-03: ⊦ m < n → m ≤ n; from specification nat