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Proof step 33 from proof from-append-add (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 34:
# 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])

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# x, n1, n2]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x, n1, n2