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Proof step 32 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 33:
# 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 inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# x, n1, n2] flags:

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x, n1, n2